Electrostatics & Electric Fields

K Warne

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Coulomb’s Law:

F =kQ1Q2

r2

FQ2

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 forceby a proportional amount.

F =kQ1Q2

r2F

Q2Q11.

F2=k2Q1Q2

r22F

Q2

2Q12.

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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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qF

E

QE =

Fq but F =kQ1Q2

r2

so E = q

kQq

r2

so E =kQ

r2

rQq

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Electric field due to multiple charges

Calculate the electric field strength at X.

XB

-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 ONLYFor FULL presentation click HERE >> ScienceCafe

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

1/2mv2 = 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.

+Q2

Q1

•r

Energy = Work done

= F x d

= kQ1Q2 x r

r2

U = kQ1Q2

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

+Q2

Q1

•r

= kQ1Q2

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 = Ed

Q Q

E = V

d where

E = electric field strength in Vm-1

V = potential difference in V and

d = 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

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Hi -

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My FULL presentations, which contain a lot more more slides and other resources, are freely

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Have a look and enjoy!

WarneScience