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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