UNIT II - Mr. Greenberg Physics - Home · Web viewPHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 25 FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th

Unit III – ELECTRIC POTENTIAL

References:PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 25

FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 25

Unit ObjectivesWhen you have completed Unit III you should be able to:

1. Define the concept of ELECTRIC POTENTIAL in terms of energy of a charged particle in an electric field and state the SI units of electric potential.

2. Given the electric potential in a region of space:a. Calculate the work done on a charged particle as it moves from one point to another.b. Applying conservation of energy, calculate the kinetic and/or potential energy of a

charged particle as it moves from one point to another.

3. State the definition of the electron-volt (eV) in terms of the motion of an electron through an electric potential difference and relate the eV to the joule.

4. Use the definition of electric potential and/or the superposition principle to find the electric potential at a point caused bya. one or more point charges.b. continuous charge distributions having planar, cylindrical, or spherical symmetry.

5. Given the electric potential as a function of distance, determine by differentiation, the electric field as a function of distance.

6. Sketch graphs of the electric field and the electric potential functions found for the charge distributions given in 4b and 5.

7. Given a sketch of the equipotential lines about a simple charge configuration, describe semiquantitatively the electric potential and the electric field about the object.

III-1


Consider a 4 kg mass close to the surface of the earth where thegravitational field is uniform.

For answers & solutions see “In-Text Fill-in Answers” beginning on page III-20

Case A: The FORCE on the mass due to the gravitational field is

= m ( ____ ) = 4 kg ( ________, ________ ) (magnitude) (direction)

= ________ , _________. (III-2 #1) (magnitude) (direction)

the force required to keep the mass at rest is _______ , _______. (III-2 #2) (magnitude) (direction)

Case B: Once it is moving, the FORCE required to move the mass up at constant speed is _______ , _______. (III-2 #3)

(magnitude) (direction)

Once it is moving, the FORCE required to move the mass downward at constantspeed is _______ , _______. (III-2 #4) (magnitude) (direction)

The expression for the WORK done by a constant FORCE acting over a

displacement is

W = ____________ (III-2 #5)

Therefore in the figure above the work done by the force while moving the 4 kg

mass from (a) to (b) through a displacement of = 1.5 m upward at constant speed is

(III-2 #6)

Definition: The GRAVITATIONAL POTENTIAL of a point in space is the WORK PER UNIT MASS required to move a test mass from infinity to the point within a gravitational field.

The difference in gravitational potential between two points in a gravitational fieldthen is the WORK PER UNIT MASS required to move a test mass between the twopoints.

Using this definition in Case B: The work done in moving the 4 kg mass 1.5 meter at constant speed upward was __________. (III-2 #7)

Thus the work done per unit mass (the difference in gravitational potential) moving the mass from point (a) to (b) was __________ _______ (III-2 #8)

(units)

III-2

Unit III – ELECTRIC POTENTIALIII-3


In electricity we have a similar idea called the ELECTRIC POTENTIAL.

Consider a (+) test charge q in a uniform electric field.

Case A: The FORCE on q due to the –field is given by

= q( ____ ); where the direction of is ________. (III-3 #1)

The force necessary to keep q at rest has a magnitude of ________ and is directed_________. (III-3 #2)

Case B: The FORCE required to move q up or down at constant speed is ________, ________. (III-3 #3) (magnitude) (direction)

The WORK done by this force as q is moved through a displacement from(a) to (b) at constant speed is

= ____________. (III-3 #4)

Definition: The ELECTRIC POTENTIAL (symbol V) at a point in an electric field is the WORK per unit charge required to move a (+) test charge from infinity to the point.

The DIFFERENCE IN ELECTRIC POTENTIAL (V) between two points in an electric field is the WORK required to move a (+) test charge between the two points PER UNIT CHARGE.

Using this definition in Case B: Let: q = +2 µC and = 25 N/C, downward. The workdone moving q upward 1 cm at constant speed is ____________. (III-3 #5)

Thus the work done per unit charge moving q from (a) to (b) is _____________. (III-3 #6)

The ELECTRIC POTENTIAL DIFFERENCE between points (a) and (b) is

__________ _____. (III-3 #7) (units)

In Unit I we found that the general expression for the work done moving a charge q from apoint (a) to a point (b) in an electric field was given by:

III-4


From the definition of potential difference and the expression for work above the potentialdifference between points (a) and (b) in an electric field is, therefore,

where V is the symbol we will use for electric potential and ∆V for electric potential difference.

Note that if the work is done in the way described in Case B then the work results in a change in electric potential energy ∆U and we can also write

Turning this expression around we find that if a charge q moves through a difference in potential ∆V, then the work done on it and the resulting change in its potental energy will be

If the work required to move a +2 C charge from infinity to a point P is 10 J, the electric potential at P is,

________ _____. (III-4 #1) (units)

The unit for electric potential is the JOULE per COULOMB. In the MKSA system of units a J/C is defined as a VOLT. That is,

Suppose an electron is in a region where there is an –field. If the electron is released, work

will be done on it by the –field causing it to gain kinetic energy and lose potential energy. Imagine two points in the field, A and B, where the difference in electric potential between them, ∆VAB, is +1 volt. If the electron is released at point A, and accelerates toward B, the loss in its potential energy will be,

∆UA->B = qe∆VAB = _________ _____ (III-4 #2) (units)

and hence its gain in kinetic energy, ∆K will be

∆K = = _________ _____. (III-4 #3) (units)

This amount of energy has a special name: the ELECTRON-VOLT (abbreviation: eV).

Definition - The amount of energy gained by an electron when accelerated through a difference in potential of 1 volt

III-5


is defined as 1 electron-volt.

III-6


Therefore, according to this definition

1 eV = ___________ Joules. (III-5 #1)

If an electron is accelerated through a difference in potential of 20 volts its gain in K will be

_______ eV or ___________ J. (III-5 #2)

A 100 MeV proton is a proton that has an energy equivalent to an electron that has been accelerated through a ∆V of _______ volts and has an energy of _______ J. (III-5 #3)

The drawing below shows three possibilities for the electric potentials at two points, A and B, in a uniform electric field. Answer the following:

a) In each case, what is the direction of the electric field? (III-5 #4)

b) If there is a (+) charge between A and B, What is the direction of the electric force on the charge? (III-5 #5)

c) Will the charge have a higher U at point A or point B? (III-5 #6)

d) In which case, if any, is the most work done on the charge by the electric force when the charge is moved from A to B? (III-5 #7)

At this point try problems 1 through 7 at the back of this unit.

Back in Unit I we derived an expression for the work done moving a point charge q from a distance RA to a distance RB from another point charge Q. This expression was

From this we can derive the electric potential about a point charge Q. If we move q from infinity to a point that is a distance R from Q, the work done will be

The work done per unit charge or the difference in potential V is then

III-7


But V∞ = 0 since here too we define the reference level for electric potential as being zero when R = . Thus, the electric potential at a distance R from a point charge Q is given by

The electric potential of a bunch of point charges at a point in space is just the sum of the potentials due to each charge at the point. That is, the potential at a point due to n point charges is

Section 25.3 in Serway covers this material. Read over Example 25.3 on page 778 then complete the following.

In the figure at the right Find the electric potential at a point P

located at the center of the square.

Solution: The distance from any charge to P is ______________. (III-6 #1)

The potential at P due only to q1 is

= __________ ______ . (III-6 #2) (units)

… due only to q2 is

V2 = __________ ______ . (III-6 #3) (units)


V3 = __________ ______ . (III-6 #4) (units)


V4 = __________ ______ . (III-6 #5) (units)

Now the total potential at P is

III-8


Therefore

Vp = __________ ______ . (III-7 #1) (units)

What four point charges, all of the same magnitude, would you place at the corners of the square (one charge at each corner), so that both the -field and the electric potential would be zero at point P? (III-7 #2)

At this point try problems 8, 9 and 10 at the back of this unit.

POTENTIAL DUE TO CONTINUOUS CHARGE DISTRIBUTIONSSuppose that instead of a bunch of point charges we have a continuous charge distribution as in the sketch at the right. The contribution to the potential at the point P due to a teensy weensy hunk of stuff having charge dQ is

.

Relative to a reference level at infinity, the total potential at P due to all the teensy weensy hunks of charge is just the sum of the potential due to each hunk. That is,

Let’s apply this to several simple cases.

The Electric Potential Of A Loop Of Wire At A Point On The Axis Of The Wire.

Before beginning this section, go back and review how the integral for calculating the for a loop was set up on page II-5 and II-6.

Consider the loop drawn at the right having a total charge of +Q distributed uniformly around its circumference. The charge per unit length,, for the loop is given by

(III-7 #3)

and since the charge is uniformly distributed, the charge dQ on the teensy weensy hunk of length ds is given by

dQ = (______)(_______) (III-7 #4)

III-9


On page III-7 we stated that the potential at a point P a distance R from a t.w. hunk of charge dQ is given by

Write this in terms of , ds, and the distance from the element ds to point P. Put all constants out in front of the integral sign.

(III-8 #1)

Since we’re integrating over all the little elements of length ds, the limits of integration are from s = _________ to s = _________. (III-8 #2)

Completing the integration we get

= ______________. (III-8 #3)

As a last step put in terms of the distance along the axis of the loop, x, the radius of the loop, R, and the total charge on the loop Q.

= ______________ ‘cause = _________ and D = _________ (III-8 #4)

This relation was derived for a positively charged loop. If Q were (–), would the equation for

be any different? If so, how would it be different?

____________________________________________________________________ (III-8 #5)

Just for fun, show that the potential at the center of the loop is given by

V (at center) = kQ/R (Hint: what is x equal to if point P is at the center?) (III-8 #6)

Show that as x becomes >> R the loop begins to look like a point charge with the potential function reducing to:

(Hint: if x >> R, then R can be neglected) (III-8 #7)

In the above example we found the on-axis potential function for a charged loop by adding up the potentials due to each teensy weensy hunk of the loop. Now let’s illustrate another way of finding the potential of the loop if we know its electric field.

Referring to page II-6, the electric field, , of a charged loop along the axis of the loop is

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If we move along the axis a teensy weensy distance toward the loop moving against the of the loop the t.w, change in electric potential is

To find Vp we must integrate from our reference level at s = 0 (x = ) to s = sp (x = xp):

To integrate we must have the integral in terms of 1 variable. Since we want the potential in terms of the distance from the loop, let’s use x as our variable. Since Thus

or in terms of the total charge on the loop Q:

This should be what you got for on page III-8 where x = xp.

Recall this integration technique:

Let

Hence

III-11


The Electric Potential Of A Uniformly Charged Disk At A Point On Its Axis.Again before beginning this section, it would be useful if you would review the set up of the integral for calculating the

–field for a disk on pages II-9 through II-11.

Consider the disk sketched at the right having a total charge Q distributed uniformly over its surface. The charge per unit area, , of the disk is

(III-10 #1)

If we imagine the super thin loop of radius r the charge dQ on this loop in terms of and r is

dQ = (charge/area)(area of loop) = (________)(________) (III-10 #2)

Previously we found that the electric potential due to a loop having charge Q at a point P a distance x along the axis of the loop was

Thus the contribution to the total potential at P due to the loop of radius r containing charge dQsketched above is

To find the total potential at P we must add up (integrate) the contributions due to all the super thin loops with radii varying from zero to _______. (III-10 #3)

Write the integral in terms of , r, x and any other necessary constants, writing all constants in front. Include the limits of integration.

(III-10 #4)

Completing the integration:

(III-10 #5)

III-12


Now let’s derive the expression for the potential of a uniformly charged disk using as we did for the loop.

The electric field of a charged disk at a point P along an axis through its center derived on page II-10 was

If we move a teensy weensy distance toward the disk against the –field the resulting change in poential is

By the same argument used for the loop on page III-9, ds = – dx and thus

This should be what you got on p. III-10

III-13


Before discussing equipotential surfaces let’s show that the potential associated with every

point within or on the surface of a solid conductor sitting in an -field is the same.

Suppose the sketch at the right represents a blob of conducting material that is sitting in an electric field. The external field causes the charges on the blob to separate. No matter what the direction of the external -field it will always be perpendicular to the surface of the

blob (see #2 page II-20). Suppose is the field at a point P on the

surface of the blob and is a teensy weensy displacement along the

surface of the blob. The change in potential due to this displacement , is

Thus, if we move along the surface of the conductor the potential doesn’t change and therefore is the same everywhere on the surface.

Suppose represents a t.w. displacement inside the conducting blob. Since the -fieldinside the conducting blob is zero (see page II-20 #1 and problem III on page II-21 and solution beginning on page 22). Thus the change in potential due to this displacement inside the blob is

This tells us then that all points inside the blob are at the same potential.

Now suppose is a t.w. displacement from a point inside to a point on the surface. Again

the change in potential due to this displacement is zero since the -field inside is zero. Thus all points inside and on the surface must be at the same potential since the change in potential in moving from an interior point to the surface is zero. Notice that it doesn’t matter if the blob is hollow or solid.

Hence we conclude that the entire conducting blob is at the same potential. (Pretty neat proof-huh?)

Let’s apply this idea to the following example:

Show that the potential graphed as a function of distance from the center of a conducting sphere of radius R and charge Q is correctly represented by the graph at the right, where Vo is the potential at the surface of the sphere.

Solution:In Unit II we found that the -field for a charged conducting sphere of radius R was

III-14


for all values of r ≥ R.

III-15


Thus the potential at any r ≥ R is given by

where here again ds = – dr since s is being measured from toward the sphere and r is being measured from the center of the sphere outward. Thus

So for all r > R the potential of a charged conducting sphere is proportional to 1/r and when

r = R the potential at the surface of the sphere is .

Since the potential inside a conducting blob (in this case the blob is a sphere) is the same as the potential at the surface then the potential is Vo throughout the sphere and decreases as 1/r outside the sphere. (Notice for all r > R the potential is the same as that for a point charge Q located at the center of the sphere.)

It should be stressed that the potential is constant throughout a CONDUCTING volume. Do not extrapolate this as true for non-conductors.

At this point try problems 11 through 14 at the back of this unit.

Equipotential Surfaces

On the preceeding pages we discussed a physical (solid or hollow) conducting blob and argued that:

1) If the charges on the blob are stationary, the -field must be perpendicular to the surface of the blob.

2) If is perpendicular to the surface of the blob, the potential is the same everywhere on the surface.

3) The potential inside the blob is constant and equal to the potential at the surface.

Condition 1 and 2 define what is called an EQUIPOTENTIAL SURFACE. It can be either an actual physical surface (as with our blob) or a purely mathematical surface in space. It is a surface that is everywhere perpendicular to the -field and every point on the surface is at the same potential. Note also if 3 is true within the volume of the physical blob, this defines as equipotential volume.

III-16


Let’s illustrate the idea of an equipotential surface using a +0.11 nC point charge. The potential of this charge as a function of the distance r from the charge is (V = 0 at ):

This means that the locus of all points that are the same distance r from our charge are at the same potential. Therefore, the equipotential surfaces for a point charge are just a bunch of concentric spheres centered on the charge.

On the grid at the right, sketch and label the three equipotential surfaces for our +0.11 nC charge that have potentials of 1, 2, and 3 volts. Recalling that the -field lines for a (+) point charge are directed radially________________ draw several

-field lines on the diagram also. After you’ve drawn

them in, note that the -field lines are perpendicular to the equipotential surfaces. Realize you’ve only drawn the surfaces in cross section. They are really spheres centered on the charge. (III-14 #1)

Now for a few blanks to fill in.1. If you moved a (+) or (–) test charge from one point on the 2v surface to another point on the same surface, you would have to do __________ joules of work. (III-14 #2)

2. If a (+) charge were moved from a distant point toward the +0.11 nC charge it would be moving toward ___________ potential. (III-14 #3)

(higher, lower)

3. If a –5 nC charge were moved from any point on the 3 V surface to any point on the 1 V surface, _________ joules of work would be done on the charge. (III-14 #4)

III-17


As an additional illustration, the drawing below is what the -field (solid lines) and corresponding equipotentials (dashed lines) look like in the vicinity of two equal but opposite charges. Since the two charges are point charges, the horizontal equipotential is at ______ V.

(III-15 #1)

The ____________ charge is the negative charge. (upper,lower) (III-15 #2)

The equipotentials above the horizontal equipotentialare at a ______ potential (III-15 #3)

(+, –)

and those below are at a ______ potential. (+, –) (III-15 #4)

In preparation for the next section notice a few things about

these -field and equipotential diagrams.

1) Look at your sketch for the 0.11 nC point charge. Notice that the change in V per unit

distance ( ) is greater closer to the charge than it is further away. That is, is

greater in a region of higher values of . For example: has a greater magnitude

between m to m than between m and 1 m. In the to m interval

whereas in the to 1 m interval . See?? The

greater the change in V with respect to r the greater the magnitude of .

2) Notice that the direction of the -field is in the direction of decreasing values of electric potential. This is, I hope, as you would expect since a (+) victim charge feels a force in the

direction of the -field. As it moves in this direction it gains kinetic energy and loses potential energy. This means the potential energy/charge decreases. Therefore, since the change in potential energy/charge is the potential, the charge is moving toward a region of lower potential.

So far we know how to calculate the potential when the -field is known. Now let’s see how to

find when the potential is known.

Consider a +q charge in an -field. Suppose we move the charge a teensy weensy distance at constant speed in a direction opposite to the -field. Under these conditions the force we must apply is

. The work done in moving +q a distance is dW and it results in an increase in the potential energy dU of the charge. Thus we can write:

III-18


Now in moving +q against the direction of means we are moving it to a region of higher potential.

III-19


The resulting change in potential dV is given by:

Rearranging we get

This relation could have been arrived at by merely differentiating

but it is better to try to understand it physically in terms of work and energy.

Recall that in the derivation was taken in the opposite direction to , that is, in the direction of the greatest increase in electric potential. We can summarize by saying the -field is equal to the negative of the rate of change of potential with respect to distance and the direction of is in the direction of the steepest decrease in potential.

Example: Given that for a point charge +Q the potential is: V = kQ/r, find the magnitude and direction of its electric field.

Solution: Since the equipotential surfaces are concentric spheres about Q the direction of the greatest decrease in V is radially outward, i.e., in the direction of r. Thus ds = dr and

So as we found in Unit I, the electric field of a point charge +Q is , radially outward.

The right hand side of the expression is referred to as the potential gradient. In some texts the expression is written as

or in fancy vector calculus notation

where is just a symbol that means “differentiate V along the direction of the greatest change in V.

Don’t let this stuff scare you.I just want to expose you to it

III-20


Let’s look at an interesting way to understand the gradient. If we plot the V for a +q charge as a function of r, the distance from the charge it looks like Figure 1. Notice that this is sort of a potential “hill” where constant elevation corresponds to equipotential lines around the “hill”. Now has the meaning of the slope or steepness or gradient of the “hill”. The

-field then is the negative of this slope or gradient at any given r. The steeper the “hill” at a particular point, the stronger the –field. Notice that the direction of is toward lower values of V. Figure 2 shows the “hill” in 3 dimensions showing again that is in the direction of maximum steepness at a given r and is perpendicular to the equipotentials.

At this point try problems 15, 16 & 17 at the back of this unit.

Now for some summary problems

I. Given the parallel plates at the right having equal but opposite charges. (The plates are large in comparison to d.)

a) Sketch in several -field vectors in the region between the plates.

b) Sketch in several equipotential lines that all have the same ∆V between them. (III-17 #1)

c) The magnitude of at point A is ________ the magnitude of at point B ( > , < , = ) (III-17 #2)

d) VB is ________ VA. ∆V = VB – VA is ________. (III-17 #3)

( > , < , = ) ( +, –, zero)e) Refer to the answer to problem 3(c) in the “End of the unit problems”. Between the

plates∆V = VB – VA = _________________. (III-17 #4)

Let a and b be the perpendicular distances from the (+) plate to points A and B, respectively, and let ∆S = b – a. The approximate gradient of the potential

between points A and B is then

= __________________ (III-17 #5)

III-21


and therefore the magnitude of the -field between the plates is

= _______________.

(III-17 #6)

f) On the grid at the right sketch what the potential “hill” looks like between the plates. Let r = 0 and at the (+) plate and r = d and V = 0 at the (–) plate (i.e., its grounded). (III-18 #1)

At this point complete Worksheet III.1.

E vs r and V vs r Graphs - Suppose we want to graph the potential, V, as a function of the distance, r, from the center of a conducting shell having a charge of +Q uniformly distributed over its surface. From earlier work (Gauss’ Law, etc.) we know that inside the shell

(for r < R) = 0 and outside the shell (r > R) . The magnitude of is plotted versus r in Figure 1. Also we know that for r > R, . Now for r < R, . That is, V is not changing as we move from the surface toward the center of the shell. V is a constant but constant at what value? Note that at the surface

of the shell (when r = R) . Since V does not change as we move inward, the potential inside the shell must also be kQ/R. The graph of V vs. r is shown in Figure 2.

III-22


Given a solid non-conducting sphere of radius R having a charge +Q spread uniformly throughout its volume, graph E and V as functions of the distance r from the center of the sphere. (III-19 #1)

a. For r > R : = ____________

b. For r = R : = ____________

c. For r < R : = ____________

d. For r > R : V = ____________

e. For r = R : V = ____________

f. For r < R : V = ____________ [hint: use and your answer to a]g. Graph d, e, & f on the V vs r grid at the right then check it with (a)

above knowing that where is the slope of the V vs r graph.

At this point try problems 18 & 19 at the back of this unit.

In the summary box on page I-10 we stated that the SI unit of the electric field is the N/C.

Notice that the relation E = - dV/dr also defines another legitimate SI unit for the -field, i.e., the Volt/Meter. Therefore, a V/m is equal to a N/C.

– End Unit III –

For a, b, & c refer to prob. II page II-21 & 22.Sketch graph for a, b, & c here

III-23


In-Text Fill-in Answers

Page Fill-in # Fill-in Answer

III-2 1 = (4 kg)(9.8 m/s2, down) = 39.2 N, down

2 39.2 N, up

3 39.2 N, up

4 39.2 N, downward

5

6

7 58.8 J

8 14.7 J/kg

III-3 1

2 qE, up

3 qE, up

4

5 0.5 J

6 0.25 J/C

7 0.25 J/C

III-4 12 –1.6 x x10–19 J

3 +1.6 x x10–19 J

III-5 1 1.6 x x10–19 J

2 20 eV or 3.2 x 10-18 J

3 ∆V of 108 volts and K of 1.6 x 10-11 J

4 The is always pointing toward points of lower potential isdirected from A toward B in all 3 cases.

5 on a (+) charge is always in the direction of the , so is directed to the right in all 3 cases.

6 (+) charges have higher U at points of V since in each case point A has the higher V then the charge has the highest U at A in all 3 cases.

III-24


7 Since and V is the same in all 3 cases, WE is the same in all3 cases



III-6 1

2 +12.7 V

3 –25.5 V

4 +38.2 V

5 +25.5 V

III-7 1 +50.9 V

2 The V due to each charge is V = kQ/R and since the Vtotal at P is the sum of all 4 potentials of the charges at the corners there must be 2 (+) charges and 2 (–) charges. Since all the Rs are the same, the sum of the 4 potentials will give Vtotal at P = 0.

Since the of each charge will be equal. If the 2 (+)

charges are on opposite corners their will be = and opposite at point P and add to zero. If the (–) charges are also on opposite corners

their will cancel resulting in zero total at P.

3

4 dQ = ds

III-8 1

2 s = 0 to s = 2R

3

4

5 The expression for would be the same.

6 When P is at the center of the loop x = 0 so

III-25


7 If x >> R then . So if P is far enough away from the loop, the loop looks like a point charge of charge Q.



III-10 1

2 dQ = rdr) 3 zero to R

4

5

III-14 1 radially outward

Equipotential radii:

2 zero

III-26


3 higher

4 10-8 J or 10 nJ

III-15 1 zero

2 lower

3 positive

4 negative



III-17 1

2 = (the -field is uniform between the plates)

3 > (V decreases in the direction of the -field), VB – VA is +

4

5

6

III-18 1

III-27


III-19 1 a)

b)

c)

III-28




III-19 1 d)

e)

f) when r = 0, V0 = 3kQ/2R

Derivation of (f): Recall that Q is (+) thus is directed outward from the center of the sphere, i.e., in the +r direction. Starting with a value of r < R and integrating from r to r = R:

represents a small displacement in the direction of the integration so can be replaced

by (outward). Since and are in the same direction and for r < R, , thus

Solving for Vr:

Since

then

for r < R.

Note that this expression also gives:

for r = 0

for r = R

III-29


End of unit problems

1. Suppose you are standing in an -field. If you move in the direction of , are you moving

toward a region of higher or lower electric potential? Use to justify your answer. If you contained a net (+) charge, would your electric potential energy increase or decrease? What if you had a net (–) charge?

2. A proton and an electron are initially at rest in an -field. Explain why, when released the proton moves toward a region of lower electric potential and the electron moves toward a region of higher electric potential.

3. Refer to Unit II and write the expressions for the electric fields due to the following charge distributions:

i. at a distance R from a long uniformly charged wire having a charge/length +.

= , (magnitude) (direction)

ii. at a distance R from a large uniformly charged sheet having a charge/area –.


iii. at a point between two large parallel uniformly charged sheets having equal butopposite charge/area .


a. Use the integral to find the potential difference between two points, A and B, having perpendicular distances a and b, respectively, from the wire in (i)

where a > b. Which point is at the higher potential?

[Ans: , point B]

b. Do the same for (ii) where a and b are perpendicular distances from the sheet.

[Ans: , point A]

c. Do the same for (iii) where a and b are perpendicular distances from the (+) sheet.

[Ans: , point B]d. In each case (i, ii, iii) does a (+) charge have a higher potential energy, point A or

point B? What if the charge were (–)?[Ans: (+) charge: (i) higher at B, (ii) higher at A, (III) higher at B (–) charge: (i) higher at A, (ii) higher at B, (III) higher at A]

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Unit III – ELECTRIC POTENTIALIII-31



4. In problem 3(a), let = +10 nC/m, a = 10 cm, and b = 5 cm. Calculate:

a. the potential difference between points A and B. [Ans: 125 V]b. the work done moving a +2 nC particle from point A to point B. [Ans: 0.25 J]c. the change in the potential energy of the +2 nC particle if, when released, it moves

from B to A. [Ans: – 0.25 J]d. the kinetic energy of the particle at B if it is fired from point A with a kinetic energy of

2 J. [Ans: 1.75 J]e. the kinetic energy of an electron in units of eV as it arrives at B after being released

from rest at point A. [ Ans: 125 eV]

5. Consider an electron between oppositely charged parallel plates as shown in the sketch at the right. The magnitude of the charge on each plate is 1.8 C. the area of each plate is 2 m² and they are 5 cm apart. Find:

a. the charge per unit area,,on each plate. [Ans: ] b. the electric field between the plates. [Ans: ~ 105 N/C from B toward A]c. the electric force on the electron when it is near plate A. near plate B. near

the midpoint between the plates. [Ans: At all points between the plates (away from

the edges) from A toward B]d. the work required to move the electron from plate B to plate A. [Ans: 8x10-16 J]e. the potential difference between the plates. [Ans: 5000 V]f. the kinetic energy of the electron in units of eV as it strikes plate B after being released

from rest at a point on plate A. [Ans: 5 keV]g. the speed of the electron as it strikes plate B after being released from a point on

plate A. [Ans: 4.2x107 m/s]

6. As you know the -field between two large parallel plates having equal but opposite charges is constant and magnitude and direction.

a. Show that the work required to move a +Q charge from the (–) plate to the (+) plateis given by W = QEd, where E is the magnitude of the -field between the plates and d is their separation.

b. Use your result in (a) and the definition of potential difference to show that the electricfield between the plates is given by E = V/d where V is the potential differencebetween the plates.

c. The ratio V/d has units of volts/meter. In many texts the units of the -field is givenin units of volts/meter. So far we have always used N/C for the units for -field.Show that a V/m is the same as a N/C.

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7. A water droplet has a mass M and has 10 excess electrons each of charge e. It is suspended motionless between oppositely charge parallel plates as shown in the sketch at the right. If the distance between the plates is D, show that the potential difference between the plates required to hold the droplet motionless is given by V = (MgD)/(10e).

8. In both Figure I and Figure II the point charge q is +1 µC and the distance from q to point A is 2 m, and to point B is 1 m.

a. In Figure I, what is the potential difference VA – VB? [Ans: -4500 V]

b. In Figure II, what is the potential difference VA – VB? [Ans: -4500 V]

c. How much work would be done moving a +10 nC charge from A to B in Figure I? in Figure II?

[Ans: Both 45 J]d. How much work would be done bringing the +10 nC

charge from infinity to point B? [Ans: 90 J]e. Is the electric potential higher or lower as we move closer to charge q?

[Ans: Higher (see problems 1 & 2)]f. Is the potential energy of a (+) test charge higher or lower as it moves closer to q?

[Ans: Higher (see problems 1 & 2)]g. How would the answers to (a) through (f) change if q = –1 µC? Ans: (a) through (d)

the signs of all answers would change. (e) through (f) answers would be lower.

9. For the situation sketched at the right, show that the potential difference VA – VB is given by:

a. Is point A or B at the higher potential? [Ans: Point A]b. Would (+) work have to be done moving an electron from A to B or from B to A?

[Ans: A to B]

c. If = 3 µC, a = 2 cm, and d = 1 cm, find the work done has a proton is moved from B to A in units of Joules and units of eV. [Ans: 0.14 pJ, 0.9 MeV]

d. If the proton is at rest at point A and is released, what will be its speed upon its arrival at point B? [Ans: 1.3x107 m/s]

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10. Two stationary charges +q are located on the y-axis as shown in the sketch at the right. A third charge –q is brought in from Infinity along the x-axis.

a. Taking V at infinity to be zero, show that V at any point alongthe x-axis due to the charges on the y-axis is given by

where x is the x-coordinate of charge –q.

b. Show that the potential energy of the movable charge is given by

c. Show that the potential Vo, at the origin when –q is at (a, 0) is

11. A nonconducting rod of length L has a uniform charge/length. A point P is a distance a from one end of the rod has shown at the right. Beginning with the expression

,

show that the electric potential at point P is given by

12. Suppose the rod in #11 was bent into a semicircle of radius R as shown at the right. If point P is at the center of the semicircle, show, starting with the same expression as in #11, that the potential at P is given by

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13. The cylindrical rod shown at the right of radius a has a charge/length of +, consisting of a charge +q distributed uniformly along its length L. Concentric with it is a thin conducting cylindrical shell of radius b also of length L. A charge of –q is placed on the outer shell.

a. Use Gauss’ Law to show that the magnitude of the electric field between the rod and shell is

for any a < r < b. What is the direction of the -field between the rod and shell?

b. Beginning with , show that the potential difference between the rod and the shell is given by

14. Two hollow concentric conducting spherical shells have radii a and b, respectively. The outer shell is grounded and the inner shell has a charge +Q spread uniformly over its surface.

a. Sketch a graph of the -field of the shells from r = 0 to r > b where r is measured from the center of the shells.

b. Recalling that “grounding” means zero potential for the object grounded, begin with

and show that the potential of the inner shell is

c. What is the potential inside the smaller shell? [Ans: ]

15. A super-huge charged sheet has a charge/area, of +0.1 µC/m2.

a. Sketch the electric field lines and equipotential surfaces above the sheet and indicatewhich equipotential surface is at the highest potential.

b. How far apart are the equipotential surfaces whose potentials differ by 50 V? [Ans: 8.8 mm]

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16. The diagram below shows a cross-section of several equipotential surfaces in a region containing an electric field having cylindrical symmetry. The difference in potential between the equipotential surfaces in the diagram is 100 volts.

a. Taking the potential to be zero at infinity, if the charge distribution causing the -field is (+) would a point on the inner or outer equi-potential be at the highest potential?

[Ans: inner]b. In which direction does the -field have the greatest magnitude? [Ans: P toward A]

the least? [Ans: P toward C] Is there a direction where the -field isroughly constant? [Ans: yep, P toward B] Generally, what is the direction of in the plane of the sketch? [Ans: outward away

from P]c. What is the difference in potential between points A and E? between D and C?

[Ans: +300 V, – 100 V]d. What is the net work done moving a +2 nC charge from A to B? from B to C to A? from E to C to D to E? [Ans: all zero]e. What is the net work done moving the +2 nC charge from A to E? from E to D? from B to A to C to D? [Ans: +0.6 J, – 0.2 J, +0.4 J]f. Answer (a) through (e) if the if the charge distribution causing the -field is (–).

17. The diagram below shows some of the equipotentials in a plane perpendicular to two parallel charged metal cylinders. The potential of each equipotential line is labeled.

a. The left-hand cylinder is charged (+). What is the sign of the charge on theother cylinder? [Ans: (+)]

b. On the drawing sketch some of the -field lines due to the charge on the

cylinders.c. What is the potential difference, VA – VB,

between points A and B? [Ans: –10 V]

d. How much work is done if a 0.5 C charge is moved along a path from point A to E and then to point D? [Ans: +15 J]

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18. The graph at the right shows how the electric potential, V, for a certain charge distribution varies with x. For each of the intervals find the x-component of the -field, then graph Ex versus x. (Ignore the behavior at the endpoints of each interval)

19. A spherically symmetric charge distribution results in an electric potential, V, given by

where r is the radius from the center of the charge and C and a are constants with dimensions of length times potential (meters.volts) and length (m), respectively. What is the magnitude and direction of the electric field?

Ans: , Direction – If C is (+), is radially outward if r > a and radially inward if r > a.

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UNIT II - Mr. Greenberg Physics - Home · Web viewPHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 25 FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th

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