Electrostatics123

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Transcript of Electrostatics123

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By:

AakankshaSimranSejal

Class: XII

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It states that the force of attraction or repulsion between any two stationary point charges is directly proportional to the product of their magnitudes of the two charges and inversely proportional to the square of the distance between them.

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Electrostatic Induction

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Electric fields

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E

+ + + + + + + + + + + + + + + + + + +

Earth

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p=qL = the electric dipole momentThe dipole moment p is defined as a vector directedfrom -ve to +ve.

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cosE2EE x2x1

icosE2E ,So net

2akq

E

a2L

acos 2

L

ia2

Lakq

2E ,So2net

rEnet

kqL

a3i

22

2Lra

23

23 2

r2L32

2L2

net )(1(1

rp

kr

pkE

qLp

rE

net;

kp

r 3

Electric Dipole

aa

E1

E1x

ij

y

For large r

E1x E cos and E1x E2x

E2

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A uniform external electric field exerts no net force on a dipole, but it does exert torque that tends to rotate the dipole in the direction of the field (align with )p

extE

1F

2F

x

Torque about the com = τ

FL sin

qEL sin pE sin rp

rE

Ep

τSo,

When the dipole rotates through the electric field does work:

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Potential Energy

Ans. The energy is minimum when aligns with

EpcospEU

Integrating,

So, U -p

E

p

E

dW τd pE sindWork done equals

The minus sign arises because the torque opposes any increase in

Setting the negative of this work equal to the change in the potential energy, we have

dsinpEdWdU

0UcospEdsinpEdWdUU 90 when0U choose We

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The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area

ΔΦ=E Δs cosθ It is a scalar quantity and it’s SI unit is

(Nm*m)/cIt has direction which is normal to the

plane.

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States the total electric flux through a closed surface (surface integral of electric field over a closed surface) is equal to 1/o times the total charge enclosed by the surface.

Mathematically

enclosed

s

qSdE 0

1.

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