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Question : 24
Total: 26
SECTION - E
(a) Derive an expression for the electric field
(b) Draw a graph of
(c) If this dipole were kept in a uniform external electric field
OR
(a) Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density
(b) An infinitely large thin plane sheet has a uniform surface charge density
Solution:
(a) Derivation of E along the axial line of dipole
(b) Graph betweenE vs r
(c) (i) Diagrams for stable and unstable equilibrium of dipole
(ii) Torque on the dipole in the two cases
Electric field atP due to charge + q is E 1 =
Electric field atP due to charge − q is E 2 =
Net electric Field atP
E 1 − E 2 =
−
=
( p = q .2 a )
Its direction is parallel to
.
(For short Dipole=
without drawing the graph)
(i) For stable Equilibrium:
is parallel to
.
(ii) For unstable equilibrium:
is antiparallel to
Torque= 0 for (i) as well as case (ii).
(Also accept,
=
×
∕ τ = p E s i n θ
OR
(a) Using Gauss's theorem to findE due to an infinite plane sheet of charge
(b) Expression for the work done to bring chargeq from infinity to r
(a)
∮ E . d s =
The electric fieldE points outwards normal to the sheet. The field lines are parallel to the Gaussian surface except for surfaces 1 and 2 .
Hence the net flux= ∮ E . d s =
=
= 2 E A
whereA is the area of each of the surface 1 and 2.
∴ ∮ E ⋅ d s =
=
= 2 E A
E =
(b)W = q
⋅
r
= q
( − E d r )
= − q
(
) d r
=
| ∞ − r |
= ( ∞ )
(b) Graph between
(c) (i) Diagrams for stable and unstable equilibrium of dipole
(ii) Torque on the dipole in the two cases
Electric field at
Electric field at
Net electric Field at
Its direction is parallel to
(For short Dipole
(i) For stable Equilibrium:
(ii) For unstable equilibrium:
Torque
(Also accept,
OR
(a) Using Gauss's theorem to find
(b) Expression for the work done to bring charge
(a)
The electric field
Hence the net flux
where
(b)
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