CBSE Class 12 Physics 2014 Delhi Set 1 Paper

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Question : 24
Total: 30
Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda.

OR
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(b) Two charged spherical conductors of radii R1 and R2 when connected by a conducting wire acquire charge q1 and q2 respectively. Find the ratio of their surface charge densities in terms of their radii.
Solution:  
(b) W‌=‌ Force ‌×‌ Displacement ‌
F‌=qE
As displacement =0, so work done is also zero.

(a)

‌E‌=‌
σ
ε0
=‌
Q
Aε0

∴‌V=Ed=‌
Qd
Aε0

‌ Capacitance, ‌‌C=‌
Q
V
=‌
ε0A
d

(b) When the two charged spherical conductors are connected by a conducting wire they acquire the same potential.
i.e., ‌
Kq1
R1
‌
=‌
Kq2
R2

⇒‌‌‌
q1
q2
=‌
R1
R2

Hence, ratio of surface charge densities,
‌‌
σ1
σ2
=‌
q1∕4πR12
q2∕4πR22

‌‌
σ1
σ2
=‌
q1R22
q2R12

‌‌
σ1
σ2
=‌
R1
R2
×
‌
R22
R12
=‌
R2
R1
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