p6

 Suppose that the area where the window is located is replaced by a wall with thick insulation. The thermal conductivity of the same area will be decreased to 0.0039 W/m/°C and the thickness will be increased to 16 cm. Determine the rate of heat transfer through this area of 2.16 m2.

Answer: 13 W

Solution:Rate = (0.0039 W/m/°C)•(2.16 m2)•(21°C - -4°C)/(0.016 m)Rate = 13 W (rounded from 2352 W)

Example Problem 1What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C.

Like any problem in physics, the solution begins by identifying known quantities and relating them to the symbols used in the relevant equation. In this problem, we know the following:

m = 450 gC = 4.18 J/g/°CTinitial = 15°CTfinal = 85°C

We wish to determine the value of Q - the quantity of heat. To do so, we would use the equation Q = m•C•ΔT. Them and the C are known; the ΔT can be determined from the initial and final temperature.

T = Tfinal - Tinitial = 85°C - 15°C = 70.°C

With three of the four quantities of the relevant equation known, we can substitute and solve for Q.

Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C)Q = 131670 JQ = 1.3x105 J = 130 kJ (rounded to two significant digits)

Example Problem 2A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.

Compared to the previous problem, this is a much more difficult problem. In fact, this problem is like two problems in one. At the center of the problem-solving strategy is the recognition that the quantity of heat lost by the water (Qwater) equals the quantity of heat gained by the metal (Qmetal). Since the m, C and ΔT values of the water are known, the Qwater can be calculated. This Qwater value equals

the Qmetal value. Once the Qmetal value is known, it can be used with the m and ΔT value of the metal to calculate the Qmetal. Use of this strategy leads to the following solution:

Part 1: Determine the Heat Lost by the Water

Given:

m = 50.0 gC = 4.18 J/g/°C Tinitial = 88.6°C Tfinal = 87.1°CΔT = -1.5°C (Tfinal - Tinitial)

Solve for Qwater:

Qwater = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)Qwater = -313.5 J (unrounded)(The - sign indicates that heat is lost by the water)

Part 2: Determine the value of Cmetal

Given:

Qmetal = 313.5 J (use a + sign since the metal is gaining heat)m = 12.9 gTinitial = 26.5°C Tfinal = 87.1°CΔT = (Tfinal - Tinitial )

Solve for Cmetal:

Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain Cmetal = Qmetal / (mmetal•ΔTmetal)

Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9 g)•(60.6°C)]Cmetal = 0.40103 J/g/°CCmetal = 0.40 J/g/°C (rounded to two significant digits)

Example Problem 3Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g.

The equation relating the mass (48.2 grams), the heat of fusion (333 J/g), and the quantity of energy (Q) is Q = m•ΔHfusion. Substitution of known values into the equation leads to the answer.

Q = m•ΔHfusion = (48.2 g)•(333 J/g)Q = 16050.6 JQ = 1.61 x 104 J = 16.1 kJ (rounded to three significant digits)

Example Problem 3 involves a rather straightforward, plug-and-chug type calculation. Now we will try Example Problem 4, which will require a significant deeper level of analysis.

Example Problem 4What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g.

In this problem, the ice is melting and the liquid water is cooling down. Energy is being transferred from the liquid to the solid. To melt the solid ice, 333 J of energy must be transferred for every gram of ice. This transfer of energy from the liquid water to the ice will cool the liquid down. But the liquid can only cool as low as 0°C - the freezing point of the water. At this temperature the liquid will begin to solidify (freeze) and the ice will not completely melt.

We know the following about the ice and the liquid water:

Given Info about Ice:

m = 50.0 gΔHfusion = 333 J/g

Given Info about Liquid Water:

C = 4.18 J/g/°C Tinitial = 26.5°C Tfinal = 0.0°CΔT = -26.5°C (Tfinal - Tinitial )

The energy gained by the ice is equal to the energy lost from the water.

Qice = -Qliquid water

The - sign indicates that the one object gains energy and the other object loses energy. We can calculate the left side of the above equation as follows:

Qice = m•ΔHfusion = (50.0 g)•(333 J/g)Qice = 16650 J

Now we can set the right side of the equation equal to m•C•ΔT and begin to substitute in known values of C and ΔT in order to solve for the mass of the liquid water. The solution is:

16650 J = -Qliquid water

16650 J = -mliquid water•Cliquid water•ΔTliquid water

16650 J = -mliquid water•(4.18 J/g/°C)•(-26.5°C)16650 J = -mliquid water•(-110.77 J/°C)mliquid water = -(16650 J)/(-110.77 J/°C)mliquid water = 150.311 gmliquid water = 1.50x102 g (rounded to three significant digits)

Jake grabs a can of soda from the closet and pours it over ice in a cup. Determine the amount of heat lost by the room temperature soda as it melts 61.9 g of ice (ΔHfusion = 333 J/g).

Answer: 20.6 kJ

Use the equation Q = m•ΔHfusion where m=61.9 g and ΔHfusion=333 J/g. Conversion to kiloJoule is of course optional.

An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styrofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc.

Answer:0.38 J/g/°C

The water warms up and the energy it gains is equal to the energy lost by the metal. The quantity of energy gained by the water can be calculated as

Qwater = m•Cwater•ΔT = (50.0 g)•(4.18 J/g/°C)•(28.1°C-27.0°C) = 229.9 J

Now this 229.9 J is equal to the -Qmetal. The specific heat capacity of the metal can be calculated by setting -229.9 J equal to m•C•ΔT.

C = -229.9 J/(11.98 g)/(28.1°C - 78.4°C) = 0.382 J/g/°C

 The heat of sublimation (ΔHsublimation) of dry ice (solid carbon dioxide) is 570 J/g. Determine the amount of heat required to turn a 5.0-pound bag of dry ice into gaseous carbon dioxide. (Given: 1.00 kg = 2.20 lb)

Answer: 1300 kJ (rounded from 1295 kJ)

mdry ice = 5.0 lb•(1.00 kg/2.2 lb) = 2.2727 kg

Now that the mass of dry ice is known, the Q value can be determined. Again, attention must be given to units. Since the mass is known in kilogram, it would be useful to express the heat of sublimation in kJ/kg. So 570 J/g is equivalent to 570 kJ/kg. And so the answer is calculated as

Q = mdry ice • ΔHsublimation-dry ice

Q = (2.2727 kg)•(570 kJ/kg) = 1295 kJ

Q = ~1300 kg (rounded to two significant digits)

Determine the amount of heat required to increase the temperature of a 3.82-gram sample of solid para-dichlorobenzene from 24°C to its liquid state at 75°C. Para-dichlorobenzene has a melting point of 54°C, a heat of fusion of 124 J/g and specific heat capacities of 1.01 J/g/°C (solid state) and 1.19 J/g/°C (liquid state).

Answer: 680 J (rounded from 684.9 J)

This problem requires three steps - calculating the Q1 for raising the temperature of para-dichlorobenzene (abbreviated as PDCB for the remainder of the problem) to 54°C (the melting point), calculating the Q2for melting the PDCB, and calculating the Q3 for raising the temperature of the liquid PDCB to 75°C.

Q1 =(3.82 g)•(1.01 J/g/°C)•(54°C-24°C) = 115.7 J

Q2 =(3.82 g)•(124 J/g) = 473.7 J

Q3 =(3.82 g)•(1.19 J/g/°C)•(75°C-54°C) = 95.5 J

Qtotal = Q1 + Q2 + Q3 = 684.9 J

 Calculate the pressure created by ocean water (density = 1.025 g/cm3), at a depth of 11.0 Km.

Given the density of ocean water is uniform at 1.025 g/cm3 = 1025 kg/m3, the weight of water is 1.005 x 104 N/m3 . The pressure of water at this depth (the weight per square meter) would be (1.005 x 104 N/m3)(11.0 x 103 m) = 1.105 x 108 N/m2 = 1.105 x 105 kPa (about 16000 psi)

 A hot-air balloon consists of a basket and a 2.18x103 m3 envelope having a combined weight of 2.45 kN. What should be the temperature of the air used to inflate the envelope to provide a net lift of 2.67 kN? The surrounding air is 20ºC, has a weight of 11.41 N/m3, has an average molecular mass of 0.028 kg/mole, and is at a pressure of 1.0 atm.

The lifting force required = 2.45 kN + 2.67 kN = 5120 N weight of 2.18x103 m3 air at 20ºC = 11.41 N/m3 * 2.18x103 m3 = 24874 N Archimedes' principle informs us of buoyancy: Lifting force = [weight of 2.18x103 m3 air at 20ºC] - [weight of 2.18x103 m3 hot air] Substituting calculated values, 

5120 N = 24874 N - [weight of 2.18x103 m3 hot air]weight of 2.18x103 m3 hot air = 19754 NFrom the ideal gas law, PV is proportional to WT Since P and V are constant,W1T1 = W2T2

or (W1 / W2)T1 = T2

 (24874 N / 19754 N)(293 K) = 369 K 365 K = 96ºC

Given the resistor arrangement shown below, prove that the relation between R1 and R2 must be R2=1.618R1 in order that the resistance of the system R be equal to R2.

R=R1+R1//R2=R1+R1R2/(R1+R2). The condition is R=R2, so we do R2=R1+R1R2/(R1+R2). R1R2+R2

2=R12+R1R2+R1R2. 

0= R22 - R1R2 - R1

2.

This is a second degree equation in R2, whose solution is R2= R1(1/2 ±  )

Disregarding the minus sign we get R2 = 1.618R1.

The circuit below is called a Wheatstone bridge. It is used for measuring resistance. Show that when the current through the galvanometer G is zero, then R1 = R2(R3/R4). Thus, if we know R2 and the ratio (R3/R4), we can obtain the resistance R1.

IG = 0 means points C and D are at the same potential and that I1 = I2, I3 = I4.

By Ohm’s law VAC = I1R1, VAD = I3R3, VCB = I2R2, VDB = I4R4.

As VAC = VAD and VCB = VDB, then I1R1= I3R3,I2R2= I4R4.

As I1 = I2 and I3 = I4, then doing I1R1/ I2R2=I3R3/I4R4 we get R1= R2(R3/R4).

Two small similar spheres with the same negative charge and 40 g of mass each are hanging at rest as indicated in the figure. The length of each string is 20 cm and the angle ß is 6º. Calculate how many electrons are needed to add to each sphere to obtain this situation.

The spheres distance is 2a and a is obtained from sin6º = a/0.2m, this is a = 0.021 m. The distance r between the spheres is then 0.042 m.

On the left sphere's free body diagram, where T is the string tension, Fs is the repulsed force the right sphere exerts over the left one and mg is the sphere weight. Then we have:

1) Sum of Fx = Tsinß - Fs = 02) Sum of Fy = Tcosß - mg = 0 or T = mg/cosß, which replaced in  1) we get: Fs = (mg/cosß)sinß = mgtgß = (40x10-3 kg)(9.8  m/s2)tan6º = 4.12x10-2 N.

The charge q on each sphere is calculated using Coulomb's law,

Fs = ke q2 /r2.

Replacing r for 0.042 m and Fs for 4.12x10-2 N and solving for q2, q2 = Fs r2/ke = (4.12x10-2 N)( 0.042 m)2/9x109 Nm2/C2. Then q = 8.98x10-8 C.

The number of electrons needed to add to each sphere is 8.98x10-8 C/1.602x10-19C = 5.6x1011 electrons.

Assume three point charges located at the vertex of a right triangle as shown in the figure, where q1 = -80µC, q2 = 50µC and q3 = 70µC, distance AC = 30 cm,

distance AB = 40 cm. Calculate the force over the charge q3 due to the charges q1 and q2.

The force directions coincide with the lines joining each pair of point charges. The force that q1 exerts over q3, F31 is of attraction. The force that q2 exerts over q3, F32 is repulsive. This is shown in the figure where the forces F31 and F32 are vectorially added to obtain F3. The distance between q3 and q1 is obtained from

(CB)2 = (AC)2 + (AB)2 = (0.3 m)2 + (0.4 m)2 

from where CB = 0.5 m

The magnitudes of such forces are:

F31 = [(9x109 Nm2 /C2)(80x10-6 C)(70x10-6 C)]/(0.5 m)2 = 201.6 N

F32 = [(9x109 Nm2/C2)(50x10-6 C)(70x10-6 C)]/ (0.3 m)2= 350 N

It is convenient to use coordinates axis as indicated, placing the origin on the charge where we need to calculate the resultant force, in this case on q3.

The resultant force is F3 = F31 + F32 . Then in terms of components x and y:F3x = F31x + F32x.F3y = F31y + F32y.

F31x = F31cosß = (201.6 N)(0.4/0.5) = 161.3 N ; F31y = - F31senß =-(201.6 N)(0.3/0.5) = -121 NF32 = 0 ; F32y = F32 = 350 N.F3x= 161.3 N + 0 = 161.3 N ; F3y = -121 N + 350 N = 229 N. 

The magnitude of the resultant force F3 is obtained from (F3)2 = (F3x)2 + (F3y)2,

giving F3 = 280 N. The angle this force makes with the x axis is calculated using tgµ = F3y/ F3x= 229/161.3 = 1.42, then µ = arctan1.42 = 54.8º.

A rapidly spinning paddle wheel raises the temperature of 200mL of water from 21 degrees Celsius to 25 degrees. How much a) work is done and b) heat is transferred in this process?

Solution:

In this problem the work is done by the friction force. All the work will go to increase the internal energy of the water, which can be calculated as:

Where   - specific heat of water,  .

There are no heat transfer in the process.

The temperature of a body is increased from -173 C to 357 C. What is the ratio of energies emitted by the body per second in these two cases?

Solution:

In this problem we need to use the Stefan–Boltzmann law: the amount of thermal radiation emitted per second by a (black) body is directly proportional to the fourth power of its absolute temperature:

where   is a (Stefan-Boltzmann) constant. It is important that in the above expression the temperature is the absolute temperature (measured in Kelvin).

It means that

We convert the temperatures of the body into Kelvin:

Now we can find the ratio of the energies emitted by the body per second:

How many btu are needed to change 10 pounds of ice at 5 degree Fahrenheit to steam at 250 degree Fahrenheit?

Solution:

The first step: we convert the temperatures given in Fahrenheit into the temperatures in Celsius:

5 degree Fahrenheit = -15 C

250 degree Fahrenheit = 121 C

Then we can see that initially we have m=10 pounds = 4.5 kg of ice and finally we have m = 4.5 kg of vapor. Therefore the whole process can be considered as the combination of the following processes:

(1) heating the ice from -15 C to 0 C;

(2) melting of ice at fixed temperature 0 C. After this process the ice becomes water.

(3) heating the water from 0 C to 100 C.

(4) vaporization of water at fixed temperature100 C. After this process the water becomes vapor.

(5) heating the vapor from 100 C to 121 C.

Process (1): the heat required to change the temperature of the ice is determined by the specific heat of ice:

The specific heat of ice is 2060 J/kg C. Then

Process (2): the heat required to melt the ice is determined by the specific latent heat of fusion:

The specific latent heat of fusion is 334000 J/kg. Then

Process (3): the heat required to change the temperature of the water is determined by the specific heat of water:

The specific heat of water is 4186 J/kg C. Then

Process (4): the heat required to vaporize the water is determined by the specific latent heat of vaporization:

The specific latent heat of vaporization is 2260000 J/kg. Then

Process (5): the heat required to change the temperature of the vapor is determined by the specific heat of vapor:

The specific heat of vapor is 1870 J/kg C. Then

Then the total heat is

We can convert the joule in btu:

f ice has a density of  , then what is the volume of 5,000 g of ice?

Solution:

The mass of ice is given in grams. The correct SI unit of the mass is kilogram (kg). We convert the mass of ice into kg:

Now we need to use the relation between the mass of the object (  ) , its density ( 

), and its volume (  ):

From this expression we can find the volume of the ice:

An ice cube having a mass of 50 grams and an initial temperature of -10 degreesCelsius is placed in 400 grams of 40 degrees Celsius water. What is the final temperature of the mixture if the effects of the container can be neglected?

Solution:

In this problem we need to use the energy conservation law. Namely, the energy transferred from the ice cube is equal to the energy transferred to the water.

Initially we have two systems: (1) ice cube and (2) water. The systems have different initial temperatures. In the final state the temperatures of the systems are the same – thermal equilibrium.

To find the final temperature we need to write the energy conservation law: the energy transferred from the system (1) [to the system (2)] is equal to the energy transferred to the system (2) [from the system (1)].

We introduce the final temperature of the systems:  . We assume that the final temperature is greater than 0. It means that in the final state the system (1) (ice) becomes water. This is our assumption – if after the calculations we obtain that the temperature is less than 0, then we need to repeat the calculations with an assumption that the temperature is less than 0 or equals to 0.

Now we need to write the energy conservation law.

(I) energy (heat) transferred from system (1) has three contributions:

Process (1): the heat required to change the temperature of 50 g = 0.05 kg ice is determined by the specific heat of the ice:

The specific heat of ice is 2060 J/kg C, the initial temperature is (-10) and the final temperature is 0. Then the change of the temperature is 10:

Process (2): the heat required to melt the ice is determined by the specific latent heat of fusion:

The specific latent heat of fusion is 334000 J/kg. Then

Process (3): The ice now becomes the water and we need to increase the

temperature of the water from 0 to the final temperature  . The heat required to increase the temperature of the water is determined by the specific heat of the water (the mass of the water is equal to the mass of the ice):

The specific heat of the water is 4186 J/kg C. Then

Then the total heat transferred from the ice is

(II) energy (heat) transferred to the system (2) has only one contribution: We just need to decrease the temperature of the water from the initial temperature 40 degrees

to the final temperature  . The mass of the water is  .

The heat required to decrease the temperature of the water is determined by the specific heat of the water:

The specific heat of the water is 4186 J/kg C, the change of the temperature of the

water is  . Then

Then the energy conservation takes the form:

From this equation we can find the final temperature of the system:

Even if a man shows no visible perspiration he still evaporates about 500 grams of water per day from his lungs. How many calories of heat are removed by this evaporation? What is the rate of heat loss in watts due to this process?

Solution:

First we calculate the energy (heat) required to evaporate the water in SI units.

The SI unit of mass is kg. We need to convert 500 g in kg:  .

The heat required to vaporize the water is determined by the specific latent heat of vaporization:

The specific latent heat of vaporization (in SI units) is 2260000 J/kg. Then

Now we can convert the energy (expressed in joules) into calories. We need to use the relation between joules and calories:

1 joule = 0.24 calories

Then

Now we can find the rate of heat loss (in watts) – the power. The rate of heat loss can be calculated as the heat loss per unit time. The watt is a SI unit. It means that we need to calculate the heat (energy) loss (in joule) per 1 second (SI unit of time).

We know that the energy   is the energy loss per one day. To find the energy loss per 1 second we need to find the number of seconds in one day:

One day the relative humidity is 90% and the temperature is 25 degrees Celsius. How many grams of water will condense out of each cubic meter of air if the temperature drops to15 degrees Celsius? How many energy does the condensation from each cubic meter release?

Solution:

An air contains water vapor. The amount of water vapor, which a cubic meter of air can contain, cannot be more than a maximum value. We characterize the amount of

water vapor in one cubic meter of air by its mass:  . Then we can tell that the mass of the water vapor in one cubic meter of air should be less than the maximum value:

The maximum mass of the water vapor depends on the temperature:  . If the vapor has the maximum mass then such vapor is called saturated vapor.

The humidity of the air is determined as the ratio of the mass of the vapor to the maximum possible vapor mass:

We know at 25 degrees Celsius the humidity is 90 %. It means that

From this expression we can find the mass of the vapor in the air:

The temperature of the air drops to 15 degrees Celsius. At this temperature the

maximum possible mass of the vapor is  . When the temperature of the air decreases the original mass of the vapor remains the same

[  ]. If this mass is larger than the maximum

mass   then part of the vapor condenses into water. The mass of the water is

Therefore to find the mass of the water we need to find the maximum mass of the vapor.

We assume that the vapor is an ideal gas. If we introduce the maximum pressure of

the vapor (  ) then the ideal gas equation takes the form (we write this equation for volume of 1 cubic meter)

Where  - molar mass of water.

Then

The maximum pressures at 15 and 25 degrees Celsius are

Then the maximum masses of the vapor are

Now we can find the mass of the water

The second question: How many energy does the condensation from each cubic meter release?

The energy released from the condensation of the water is determined by the specific latent heat of vaporization:

The specific latent heat of vaporization (in SI units) is 2260000 J/kg. Then

A cylindrical vessel of radius 0.1 meter is filled with water to a height of 0.5 meter. It has a capillary tube 0.15 meter long and 0.0002 meter radius fixed horizontally at its bottom. Find the time in which the water level will fall to a height of 0.2 meter.

Solution:

In the present problem we need to use the Bernoulli's law and the continuity equation (liquid is incompressible).

We introduce points B and C as shown in the figure below. The velocities at these two points are

 and 

From the continuity equation we know that the product of the velocity and the cross section area is constant. We apply this equation for points C and B:

The radius of vessel at point C is  , then the cross section area at point C is

The radius of tube at point B is  , then the cross section area at point B is

Then

and

...........................................................(1)

Therefore the velocity of liquid at point C is much smaller than the velocity of the liquid at point B.

Now we can write down the Bernoulli's law for points B and C:

Since   then we can disregard the second term in the left hand side:

Then

Using the relation (1) we obtain

At the same time the velocity at point C can be expressed as the change of the height of the liquid in the vessel:

Then

Now we need to take the integral (from initial to the final moment) of both sides of the equation

Calculate the amount of energy needed to break a drop of water of diameter 0.001 meter into 1000000 droplets of equal sizes. The surface tension of water is 0.0072 N/m.

Solution:

In this problem we need to take into account only the surface energy of water drops.

The energy of a drop of water of diameter   is

Where   is the surface tension and  is the area of a spherical droplet. Then

We know the initial diameter, then we can find the initial energy

In the final state all droplets have the same diameter,  . To find this diameters we need to use the fact that the initial volume of the water is equal to the final volume:

where   is the number of droplets. Then

Then the energy in the final state is

Then

A force F is applied on a square plate of side L. If the percentage error in determination of L is 3% and that in F is 4%. what is the permissible error in pressure?

Solution:

The pressure is defined as the ratio of force and area:

where area   of a square plate of side   is  . Then

We introduce the error  in determination of the force and the error  in determination of the length. Therefore, the force is

And the length is

We know the percentage errors of the force and the length:  (4%)

and  (3%). Then

In the above expressions we should use the absolute values since the errors can be positive and negative.

Now we can express the pressure in terms of the errors of the force and the length:

We consider   and   as small parameters and use the Taylor expansion of the

above expression to find the linear terms in   and  :

Then the error in pressure is

........................................................(1)

We know that

Then the maximum error in pressure is

(it is important that we add these two corrections). Then

A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above each point and a vertical column underneath. The steel cable is 1.27 cm in diameter and is 5.75 m long before loading. The aluminum column is a hollow cylinder with an inside diameter of 16.14 cm, an outside diameter of 16.24 cm, and unloaded length of 3.25 m. When the walkway exerts a load force of 8500 N  on one of the support points, how much does the point move down?

Solution:

In this problem we need to use the Hook's law for the elastic force, expressed in terms of modulus of elasticity (Young's modulus):

where   is an elastic modulus,   is the cross-section area,   is the length of the cable

(column), and   is the change of the length of the cable (column).

The walkway with the load is in equilibrium. It means that the net force acting on the walkway is zero. There are three forces acting on the walkway: the load force, the elastic force due to cable, and the elastic force due to column. The forces are shown in the figure. Then in equilibrium we have:

..................................................(1)

The elastic forces can be found from the Hook's law:

It is important that   is the same in both expressions.

The cross-section areas are the following:

The elastic modulus can be found from the tables

Then equation (1) becomes

From this equation we can find x:

1.   a) What is the electric field of an iron nucleus at s distance of 6.00*10^-10 m from the nucleus? The atomic number of iron is 26. Assume that the nucleus may be treated as a point charge.       b) What is the electric field of a proton at a distance of 5.29*10^-11 m from the proton? (This is the radius of the electron orbit in the Bohr model for the ground state of the hydrogen atom)

2.   What is an electrical arc? How and why does it occur?

3.   A small object carrying a charge of -55 microcoulombs experiences a downward force of 6.20 x 10-9 N when placed at a certain point in an electric field.

a) What are the magnitude and direction of the electric field at this point?.b) What would be the magnitude and direction of the force acting on a copper nucleus (atomic number= 29, atomic mass= 63.5 gmol-1) placed at this same point in the electric field?

4.   A negative point charge q1 = -4 nC is on the x-axis at x = 0.6 m. A second point charge q2 is on the x-axis at x = -1.2 m. What must be the sign and magnitude of q2 for the net electric field at the origin to be a) 50 NC-1 in the +x-direction? b) 50 NC-1 in the -x-direction?

5.   A charge q1= -1 x 10-6 C is on the x-axis at x = 1.2 m, and the second charge q2 = + 2 x 10-6 C at x = 1.2 m and y = 1.6 m. a) Find the x and y-component of electric field at the origin 0. b) Find acceleration of a proton is released at the midpoint between q1 and q2.

Answer:

1.   (a)The solution to this problem invloves a simple equation from electrostatics for a point charge.

The equation is

E = kQ/r2

where

E = the electric field you want to find in NC-1 or Vm-1

k = an electostatic constant = 9 x 109 N(mC)-2 in vacuumQ = charge in C

(a) To solve the first part of the problem, you must have the charge of the iron nucleus. You must be aware that in magnitude the charge on a proton is the same as that of an electron but only the sign is opposite.

Further atomic number (Z) gives a representation for the charge of the nucleus and a neutral atom.

Here you're given

Z = 26

Therefore

Q = Ze   = (26)(1.6 x 10-19)   = 4.16 x 10-18 C

The electric field would be

E = (9 x 109)(4.16 x 10-18)/(6 x 10-10)2

   = 1.04 x 1011 NC-1

      (b)Using the same principle for the second part of the problem, you'll have

E = (9 x 109)(1.6 x 10-19)/(5.29 x 10-11)2

   = 5.15 x 1011 NC-1

2.   You have to get few basic facts clear before trying to understand my answer to your fascinating question.

The first and foremost principle is that electricity can only flow if there is a complete path for charges to flow. Normally air is a good insulator. However, there are occasions when this insulation is broken down such as with lightning that occurs naturally and arc lamps which are man made.

If charges are to flow then as we just said there must be a complete path or at least a network for the charges to pass on from one place to another. This is impossible in air unless you CREATE charges in the atmosphere by some articicial means that can conduct these and other charges. These charges can be created in the atmosphere when you by some means "TEAR" electrons off from the atoms in the atmosphere. This process is called IONIZATION and can created by various means such as supplying high electical energy as in the case of "arcing" or using radioactive sources such as radium and so on.

Here we are only interested in the former. Let's go into that in a little more detail. Normal air is neutral. By applying a high potential difference, you can ionize the atmosphere and at very high temperatures you have what's sometimes referred to as the fourth state of matter called PLASMA. Plasma is simply hot ionized gas. Charges of all kinds are floating around and it is belived that over 99% of the universe is made of plasma, our sun included!

Incidentally physicists have been trying in vain to harness this power of plasma (sun's source by nuclear fusion) by various techniques for over four decades. May be one day this source by "laser zapping" could be economically viable?

Coming back to the electical arc, have you understood the mechanism? You supply a high potential difference in between two junctions as in a carbon arc lamp, or the more common flurorescent lamp at homes and the ionization of the matter in between causes the familiar electrical arc.

Recall me telling you in the beginning that you have to understand a few principles? One more useful principle is how light is produced in general. Physicists usually explain this by turning to electron energy levels in an atom. Light is produced when an electron FALLS from a higher energy level to a lower energy level. The difference of energy between the two levels is what is emitted as light according to a famous physicist Max Planck who revolutionized physics with this suggestion first in 1900. Of course he did not have the courage to extend this and it was our dear old EINSTEIN who actually spread the concept in 1905.

The "arcing" is occuring because the ionized gas (plasma) has electons which are falling into lower energy levels.

I hope you've understood all this and do not hesitate to contact me should you have further questions on this.

When I was in school I think in Grade 7/8 my project was a simple carbon arc lamp. I was as fascinated with it then. I though you too may enjoy reading about the History of Electric Lighting Technology at the URL:

    http://sheldonbrown.com/marty_light_hist.html

Pay special attention to the Carbon Arc Lamp which was my project and another useful link would be reading about the Discovery of the Carbon Arc, The Carbon Arc Experiment That Can Be Done in the Classroom (after taking the necessary precautions) and so on at the URL:

    http://myron.sjsu.edu/caesars/CARBARC.HTM

I'm going to stop writing more now.

3.   The answer to the first part of the problem involves the use of a very important equation in electostatics.

F = Q E

where

   F = electric force in N   Q = charge in C   E = electric field in NC-1

The other principle you must know is the DIRECTION of an electric field is traditionally taken to be the direction in which a UNIT POSITIVE CHARGE would move when placed at that point in the field.

a) You're given here F and Q

Making E the subject of the equation, you'll have

E = F/Q    = 6.20 x 10-9 /(55 x 10-6)   = 1.13 x 10-4 NC-1

I've not taken the sign of the charge in the above equation to find out the magnitude of the force.

Obviously by looking at the principle above, if the charge is negative and experiences a downward force, the direction of the electric field here must be UPWARD. (negative charge attracted by the positive of the electric field, don't forget LIKE CHARGES REPEL and UNLIKE CHARGES ATTRACT).

    b) Using the same equation above and the other I gave you last time Q = Ze, you'll have

F = (29)(1.6 x 10-19)(1.13 x 10-4)   = 5.2 x 10-22 N

ACTING UPWARD at that point in the electric field.

4.   The first task for you will be drawing a straight line. Draw a rough sketch approximately to scale. On the left you should have q2 the unknown charge, in between the origin O and on the other side q1.

The formula to calculate the electric field is

E = kQ/r2

The electric field due to q1 would be

E = (9 x 109)(4 x 10-9)/(0.6)2

    = 100 NC-1 in the positive x direction.

I would recommend that you solve for practice some of my other problems on electic field where I have elaborated on these rules.

(i) In the first part of the problem you want the net electric field to be 50 NC-

1 in the +x-direction.

The formula is given by

Enet = E1 + E2

50 = 100 + E2

That is

E2 = - 50 N/C

    = kq2/r2

Therefore

- 50 = (9 x 109)[q2]/(1.2)2

giving you

q2 = -8 nC

(ii) In the second part of the problem you want the net electric field to be 50 NC-1 in the -x-direction.

The formula once again is given by

Enet = E1 + E2

-50 = 100 + E2

That is

E2 = - 150 NC-1

    = kq2/r2

Therefore

- 150 = (9 x 109)[q2]/(1.2)2

giving you

q2 = -24 nC

5.   The first step you must undertake is to draw a rough sketch approximately to scale. If you did that, you'll find that the electic field due to the first charge is towards the positive x-axis. The field due to the second

charge is along the third quadrant, making an angle of 53o below the negative x-axis.

You have to now resolve the field into the two perpendicular directions.

Considering only the second charge now. The electric field is

E2 = kq2/r2

    = (9 x 109)(2 x 10-6)/(2)2

where you've found the distance between the second charge and the origin using Pythagoras to be 2 m.

E2 = 4500 NC-1 53o below the -x axis

We can now resolve this into two components, one along the negative x-axis and the other along the negative y-axis, giving you

E2x = 4500cos53    = 2708 NC-1

and the negative y-direction

E2y = 3594 NC-1

The field due to the first charge as we said earlier will be along the positive x-axis and equal to

E1x = kq/r2

    = (9 x 109)(1 x 10-6)/(1.2)2

    = 6250 NC-1

The resultant field along the x-direction is therefore

E = 6250 - 2708    = 3542 NC-1 along the +x-direction and

E = 3594 NC-1 along the -y-direction.

b) To solve the second question you have to find the RESULTANT FORCE on a proton at the midpoint between q1 and q2.

This is done using the same principle and given by

Resultant F = F1 + F2

F1 will be vertically downward and have a magnitude given by Coulomb's law

F1 = kQq/r2

    = (9 x 109)(1 x 10-6)(1.6 x 10-19)/(0.8)2

    = 2.25 x 10-15 N

where you've taken the charge on the proton to be the standard value 1.6 x 10-19 C

and the force due to the second charge also will be vertially downward given by

F2 = kQq/r2

    = 4.5 x 10-15 N

The total will be

Fnet = 6.75 x 10-15

Therefore using Newton's second law the acceleration

a = F/m

A remote control airplane is sitting on the runway and the operator revs the engine to full thrust and then releases the brake.  The remote control plane accelerates down the runway at 1.6m/s2 and reaches it’s take off speed at 23 m/s.  How long did it take the remote control plane to take off (from the moment the brake was released)?  How far down the runway did the remote control plane travel?

Two small spheres spaced 35.0cm apart   have equal charge.  How many excess electrons must be present on each sphere if the  magnitude of the force of repulsion

between them is  ?

Solution

The Coulomb force between two equal charges (spheres) is

We know :  ,  , electrostatic

constant:  . Then we can find q :

Then we can find the number of excess electrons (since we know the charge of a

single electron  :

A point charge   is at the point  ,  , and a second point

charge   is at the point  ,  .  Find the magnitude and direction of the net electric field at the origin.

Solution

The magnitude of electric field due to charge 1 is

The magnitude of electric field due to charge 2 is

To find the net electric field we need to find the vector sum of electric field due to charge 1 and electric field due to charge 2. It is easier to work in term of components. So the x-component of the net electric field is the sum of x-components of electric field

due to charge 1 and due to charge 2, and the y-component of the net electric field is the sum of y-components of electric field due to charge 1 and due to charge 2.

The x and y components of electric field due to charge 1 are

Both components are positive since the charge is negative.

The x and y components of electric field due to charge 2 are

It is negative since the charge is positive.

Then we can find the x and y components of the net electric field:

Then we can find the magnitude of the net electric field:

The direction of electric field is determined by the angle between the vector of electric field and axis x:

What must the charge (sign and magnitude) of a particle of mass 5 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 800 N/C?

Solution

We have equilibrium. It means that the net force is 0. Then the gravitational force should be equal to electric force:

From this equation we can find the charge of the partice

The direction of electric force should be upward (since the gravitational force has downward direction). Since the direction of electric field is downward then the electric charge should be negative:

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Solution

The electric force is equal to gravitation force. So we can write:

From this equation we can find the electric field

A particle has a charge of -8.00 nC. Find the magnitude and direction of the electric field due to this particle at a point 0.5 m directly above it.

Solution

The magnitude of electric field due to point charge is given by the expression:

Since the charge is negative then the direction of electric field is downward (toward the charge).

Two particles having charges of 0.70 nC and 12 nC are separated by a distance of 2 m. At what point along the line connecting the two charges is the net electric field due to

the two charges equal to zero?   

Solution

Introduce the distance a from the charge 0.7 nC to the point P, where the net electric field is 0. The net electric field at point P is the sum of two contributions: electric field due to charge 0.7 nC and electric field due to charge 12 nC. They have opposite directions (since both charges are positive). Then the condition that the net electric field is 0 is the following:

where r=2 m is the distance between the charges. From this equation we can find the distance a :

A closed surface encloses a net charge of 10 nC . What is the net electric flux through the surface?

Solution

From Gauss's law the flux through the surface is

where 

Each square centimeter of the surface of an infinite plane sheet of paper has excess electrons.  Find the magnitude and direction of the electric field at a point 6.00 cm from the surface of the sheet, if the sheet is large enough to be treated as an infinite plane.

Solution

Electric field due to infinite plane:

Where  . The surface charge density (charge density per square meter) is:

Then the magnitude of electric field is

The direction of electric field is toward the plane (since the plane is negative charged).

A thin disk with a circular hole at its center, called an annulus has inner radius  a nd

outer radius  . The disk has a uniform positive surface charge density  . Find the total electric charge on the annulus.

Solution

By definition the total electric charge is equal to the product of the surface charge density and the area of the surface. The area is

Then the total charge is

A thin disk with a circular hole at its center has inner radius   a nd outer radius  .

The disk has a uniform positive surgace charge density  on its surface. The disk lies in the yz plane, with its center at the origin. For an arbitrary point on the x axis (the axis of the disk) find the magnitude of the electric field.

Solution

Disk can be considered as a combination of many rings. Electric field due to a ring of radius R has the following expression:

The linear charge density of a ring is  . To find the net electric field we just

need to integrate the above expression from   a nd  :

A thin disk with a circular hole at its center has inner radius   a nd outer radius  .

The disk has a uniform positive surgace charge density  on its surface. The disk lies in the yz plane, with its center at the origin.

A point particle with mass m and negative charge -q is free to move along the x axis

(but cannot move off the axis). The particle is originally placed at rest at x=0.01   and released. Find the frequency of oscillations.

Solution

From the previous problem we know the expression for electric field

At small x we have:

Then the force acting on a negative charge (-q) is

This force has a form   for which the frequency is  . Then the frequency is

Suppose that   of hydrogen is separated into its constituent protons and electrons, and that the protons are all grouped together at one point, and the electrons at another point, 3 meter distant. What is the magnitude of the force between the two groups of charge?

Solution

The first step is to find how many protons (and correspondingly electrons) there are

in   of hydrogen. To find this number we just need to divide the total mass by the mass of a single proton:

Then the charge of all protons grouped is   (where  is the proton

charge), while the charge of all electrons grouped is  . Then the magnitude of the Coulomb force is

Two small beads having positive charges 25q and q are fixed at the opposite ends of a horizontal insulating rod, extending from the origin (the location of the larger charge) to the point x = d. A third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium?

Solution

Let us assume that the position of the third charge is x (the distance from the larger charge) and the charge of the third charge is Q. Then the equilibrium condition is: the net force is 0. This condition can be written as

From this equation we can find the distance x:

roblem 14.

  Each of the protons in a particle beam has a kinetic energy of  . What are the magnitude and direction of the electric field that will stop these protons in a distance of 2 m?

Solution

To stop the proton the direction of electric field should be opposite to the direction of the beam.

To find the magnitude of electric field we need to write down the energy conservation: the initial energy ( K ) of proton should be equal to the work done by an electric field:

Where  . From this equation we can find electric field:

Two identical spheres are each attached to a rope of length 1 m and hung from a common point. Each sphere has mass 1 kg. The radius of each sphere is very small

compared to the distance between the spheres, so they may be treated as point charges. The spheres have positive charges. The spheres are in equilibrium and each rope makes an angle 30 degrees with the vertical. Find the tension of the ropes.

Solution

The system is in equilibrium. It means that the net force is 0. It means that the sum of the tension force and the gravitational force should have horizontal direction. Then we can write:

From this equation we can find the tension:

A solid metal sphere with radius 0.75 m carries a net charge of 0.13 nC. Find the magnitude of the electric field at the following locations:a) at a point 0.15 m outside the surface of the sphere. b) at a point outside the sphere, 0.1 m below the surface.

Solution

The specific property of conductor (metal) is that the electric field inside conductor (in equilibrium) is 0. From this property it is possible to obtain that the charge is distributed over the surface of conductor (not in the bulk).

In the present problem the conductor has the shape of sphere. It means that all the charge 0.13 nC will be on the surface of the sphere.

Then the electric field outside of the conductor is the same as the electric field of the point charge at the center of the sphere.

Then electric field inside of the conductor is 0.

(a) a point 0.15 m outside the surface of the sphere is at distance 0.15+0.75 = 0.9 m from the center of the sphere. Then electric field at this point is

How many excess electrons must be added to an isolated spherical conductor 27 cm in diameter to produce an electric field of 1450 N/C just outside the surface?

Solution

The electric field just outside of the spherical conductor is

where   is the radius of the spherical conductor.

We know the electric field. Then we can find the excess charge:

To find the number of excess electrons we just need to divide the excess charge by the charge of a single electron:

To study the structure of the lead nucleus, electrons are fixed at a lead target. Some of the electrons actually enter the nuclei of the target, and the deflection of these electrons is measured. The deflection is caused by the charge of the nucleus, which is distributed approximately uniformly over the spherical volume of the nucleus. A lead

nucleus has a charge of +82e (  ) and a radius of  . Find the acceleration of an electron at the following distances from the center of a nucleus.

(a) R

(b) 2R

(c) R/2

(d) 0 (at the center)

Solution

In the present problem we have a uniformly charged sphere. The electric field of such sphere has the following expression (Q is the charge of the sphere):

Then the electric force on an electron is

The acceleration is the ratio of the force and the mass of the electron:

where  .

Then

(a) at r=R:

(a) at r=2R:

(a) at r=R/2:

(a) at r=0:

An electrostatic field is given by   

Determine the work done in moving the charge 

(a) from (0,0,0) to (4,0,0) m

(b) from (4,0,0) to (4,2,0) m

(c) from (4,2,0) to (0,0,0) m along a straight path.

Solution

If the charge is moving from point A to point B then the work done by electric field is given by the expression:

In this expression the integral is calculated along any trajectory connecting points A and B.

(a) We calculate the integral along the straight line connecting initial and the final points:

(b) We calculate the integral along the straight line connecting initial and the final points:

(c) We calculate the integral along the straight line connecting initial and the final points:

A cylinder with radius r= 0.75 m and length l=0.6 m that has an infinite line of positive

charge running along its axis.  The charge per unit length on the line is  .

(a) What is the electric flux through the cylinder due to this line of charge?

(b) What is the flux through the cylinder if its radius increased to r= 0.32 m?

(c) What is the flux through the cylinder if its length increased to l= 0.8 m?

Solution

From Gauss's law the flux through the surface depends only on the charge inside of this surface.

where 

(a) The charge inside of the cylinder depends only on the length of the cylinder. It is

Then

(b) The charge inside of the cylinder depends only on the length of the cylinder, so in this case the charge is the same as in case (a). So the flux will be the same:

(c) Now the length is 0.8 m. Then

Then

Some artillery men place an old cannon over the sea level at the top of a cliff. The projectile is shooted horizontally with an initial velocity Vi. The cannon is 60 m above the sea level. The time taken since the instant the projectile is shot until its impact is heard by the artillery men, is 4.0 s. Knowing that the sound velocity in the air is 340 m/s, calculate the horizontal distance x from the impact point on the sea to the base of the cliff and the initial velocity Vi of the projectile.

ANSWERS : x = 159 m; Vi = 45.4 m/s

Placing the origin of the coordinate system in the sea along with the cliff and making x to show the horizontal distance and y for the altitude, we have:

(A) x = Vi cos t

(B) y = yi + Vi sen t - (1/2)gt2. yi is the initial vertical position.

In this case yi = 60 m, Vi is the magnitude of the initial velocity, is the angle that the initial velocity vector makes with the x axis, in this case = 0º because the shot is horizontal. So sen0º = 0 y cos0º = 1.

The time the projectile takes to reach the water is obtained doing y = 0.

y = 0 = 60m + Visen0º t - (1/2)gt2. Then,(1/2)gt2 = 60m, from where t = 3,5 seconds.

Then, the time the sound takes to travel the distance between the impact point and the cannon is 4s - 3,5s = 0,5s.

The distance between the impact point and the cannon is calculated using the known relation: distance = velocity*time = 340 m/s*0.5 s = 170 m. These 170 m form the hypotenuse of the rectangle triangle with 60 m and x as the sides, then

x2 = (170m)2 - (60m)2.

Solving x in this triangle rectangle, x = 159 meters.

The initial velocity, when the shooting ocurred, can be calculated with the equation (A)

x = Vi cos t159m = Vi *1*3,5sThen Vi = 45,4 m/s.

Find the electric field at point P on the y axis and at a distance 0.4 m above the origin. The electric field is created by the three point charges as shown below. The charge q1 = 7x10-6 C is in the origin of the coordinate system, the charge q2 = -5x10-6 C is placed on the x axis 0.3 m from the origin and the charge q3 = -3x10-

6 C is placed to the right of point P and at 0.4 m above q2.Determine also the force exerted on a charge of 3x10-6 C when placed on the point P.

SOLUTION:Let's first calculate the magnitude of the electric field at P due to the presence of each charge. Denote E1 the electric field produced by q1, E2 the electric field produced by q2 and E3 the electric field produced by q3. These electric fields are represented in the figure and their magnitudes are, using positive values for the charges:

E1 = keq1/r12 = (9x109 Nm2/C2)(7.0x10-6 C)/(0.4m)2

= 3.9x105 N/C.E2 = keq2/r2

2 = (9x109 Nm2/C2)(5.0x10-6 C)/(0.5m)2

= 1.8x105 N/C.E3 = keq3/r3

2 = (9x109 Nm2/C2)(3.0x10-6 C)/(0.3m)2

= 3.0x105 N/C.

Vector E1 has no x component, just y component (upward). Vector E2 has a x component given by E2cosß = (3/5)E2 and a negative y component given by -E2 senß = -(4/5)E2. Vector E3has no y component, just x component (to the right).

The resultant vector E we are looking for is the vectorial sum of this three vectors, E = E1 + E2 + E3. The vectors E1, E2 and E3are denoted using unitary vectors i and j. Then we can obtain their sum analytically.

E1 = 3.9x105 j N/C.

E2 = (1.1x105 i - 1.4x105 j )N/C.E3 = 3.0x105 i N/C.

The electric field E at P is then:

E = (4.1x105 i + 2.5x105 j) N/C.

The force on a charge of 3x10-6 C when is placed at the point P is obtained using F = Eq, with q = 3x10-6 C, F = (1.23 i + 0.75 j)N.This force has of course the same direction of the electric field E, because the charge q is positive.

In the figure a thin rod of length L and a total charge Q uniformly distributed is placed along the x axis. Find the magnitude of the electric field at point A located a distance d from the left side of the rod.

This problem can be solved assuming the rod is made of point charges of magnitude dq = (Q/L)dx. Q/L is the charge per unit length. We can use E = Kq/x2 in differential form, dE = Kdq/x2. dE is the contribution to the field of dq at a distance x from A. Then we add the contributions of all the electric point charges using limits of integration from d to d + L: