G11 Electrostatics & Electric Fields

Electrostatics & Electric Fields

K Warne

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Coulomb’s Law:Coulomb’s Law:

F =

kQ1Q

2

r2

F

Q2

The electrostatic force between two point charges is directly proportional to

the product of the charges and inversely proportional to the square of the

distance between them.

Q1

k = coulomb’s constant

= 9 x 109 N. m2.C-2

Calculate the force between an electron and a proton if the distance between them is 1nm. (e- = -1.6x10-19 )

F = kQ1Q2/r2

= (9x109)(-1.6x10-19 )(1.6x10-19)/(1x10-9)2

= -2.304x10-10N

Increasing the

charge on any

one of the

spheres will

increase the

force by a

proportional

amount.

Increasing the

charge on any

one of the

spheres will

increase the

force by a

proportional

amount.

F =

kQ1Q

2

r2

F

Q2

Q1

1.

F2=

k2Q1Q

2

r2

2F

Q2

2Q12.

F2=2F

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7.1 ELECTRIC FIELD

A region ____________ in which a charge will experience a “_________” or electrostatic

_______________________.

+ -+

+ -

ELECTRIC FIELD LINE:

A line drawn in such a way that at at any point on the line, a

small ___________ charge will experience a ___________ in the

direction of the ______________ to that line.

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7.2 ELECTRIC FIELD PATTERNS:

- +

++

VERY SMALL POINT CHARGES NEAR ONE ANOTHER

- +

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+ + + + + + + + + + +

- - - - - - - - - - -

Between oppositely charged plates

Field is _______________ the oppositely charged plates.

(Force experienced by a charge placed anywhere between the plates is

___________________)

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ELECTRIC FIELD STRENGTH (E)

For an electric field E = /

where E = ______________

strength in ______

F = _______ in___

Q = _________ in ___

NB. Electric field strength is a ___________quantity

(direction: _____________ to ____________)

+ F

E

Eg: What force would be experienced by an electron in an electric field of 1 x 10-6 NC-1?

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q F

E

Q E =

Fq

but F =

kQ1Q

2

r2

so E =q

kQq

r2

so E = kQ

r2

r

Qq

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Electric field due to multiple charges

Calculate the electric field strength at X.

X

B

-2C

A

+2C

1cm2cm

Field due to A:

E = kQ1

/r2

=(9x109)(2x10-6)/(2x10-2)2

= 4.5 x 107 N.C-1 away from A

Field due to B:

E = kQ1

/r2

=(9x109)(-2x10-6)/(1x10-2)2

= -1.8x108 N.C-1

= 1.8x108 N.C-1 towards B

E = E1

+ E 2

= 4.5 x 107 + ( 1.8x108 ) = 2.25x108 N.C-1

E = 2.25x108 N.C-1 TOWARDS B (AWAY FROM A) SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY

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Work done W = F x d

but F = QE (Def of E)

W = QEd

If the charge is pushed to the left the work done on

the charge is: W = QEd

If the charge is now released, it moves spontaneously to the right,because the field does work on the charge:

W = QEd

Kinetic energy gained = work done

1/2

mv2 = QEd => v = √2QEd/m

Consider applying a force F needed to move a charge from A to B. The charge moves a distance d. The size of

the charge is Q.

7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD

+ d -

+ -

+ -

+ B A -

+ -

+ -

+ -

+ -

+ -

+ -

+ -

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Question A proton is accelerated by an electric field of 1.5x106 N.C-

1 over a distance of 2 nm. The mass of a proton is 1.7 x 10-27 kg

Calculate the final velocity attained by the proton if it started from rest.

E = 1.5x106 N.C-1 d = 2 nm = 2 x 10-9 m m = 1.7 x 10-27 kg

Q = 1.6 x 10-19 C v = ?

Formula?

v = √2QEd/m

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• A positive charge moves spontaneously

… ............................................... of the field.

• A …………………………….. moves spontaneously in a direction …………… to

that of the electric field.

• Thus, at any point in an electric field an electric charge possesses

…………………… (………..)

• Where free to move, it will ………………..

• It will therefore gain ………. as it loses …….

ie. …………… = ………………. (ignoring air friction).

……………………… = ……………………

7.5 WORK, Ep AND Ek IN ELECTRIC FIELDS

+ -

+

-

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Energy in a radial field

Work must be done on Q2 to move it across a distance r from Q1.

If released Q2 must therefore have the potential to move away again with that same energy.

Q2 therefore has Potential energy U at a distance r from Q1.

+Q

2

Q1

•r

Energy = Work done

= F x d

= kQ1

Q2

x r

r2

U = kQ1

Q2

r

A positive test charge is at a ……………….. …………………. Ep

at B

and at a ………… at A.

There is thus a potential difference (V) between B and A.

POTENTIAL DIFFERENCE (V)

+

B A

Potential difference =

Unit: ………………………………….

The potential difference between two points in an electric field is the

………………………………………………………. in moving the charge from the one point to the other.

Define the electric potential at a point as the …………………………… per ………………, i.e. the potential energy

…………………………. would have if it were placed …………………...

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Electric Potential

+Q

2

Q1

•r

= kQ1

Q2

r

Q2

= kQ1

r

Charge at that point

+ -

+ -

+ -

+ -

+ -

+ -

+ -

+ -

ELECTRIC FIELD BETWEEN 2 PARALLEL PLATES

If plates are s apart and p.d across plates is V, then

V = W = QEd = EdQ QE = V

d whereE = electric field strength in Vm-1

V = potential difference in V andd = distance apart in m

NOTE: V.m-1 = N.C-1

[V.m-1 = J.C-1.m-1 = N.m.C-1m-1 =N.C-1]

V

d

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Milikan’s Experiment

• These droplets are charged as they are forced out of the nozzle.

• Milikan used a microscope to observe the oil droplets between the plates.

• As the plate voltage increases some of the drops fall more and more slowly until the drops stop moving.

• At this point the electric force is equal to the weight of the oil droplet.

V = Ed = (F/q).d .: F = q.V/d Felec

= Fg

qV = mg

d

q=Droplet Charge V= Holding Voltage  d= Distance between plates  m=droplet mass g= Acceleration due to gravity.

• From his experiments Milikan determined that the charge on an electron was 1.6×10-19 C.

• By timing how long it takes for a droplet to fall with the plates switched off he could calculate the mass of the droplet.

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Hi -

This is a SAMPLE presentation only.

My FULL presentations, which contain loads more slides and other resources, are freely available on my resource sharing website:

www.sciencecafe.org.za

(paste into your browser if link above does not work)Have a look and enjoy!

Keith Warne