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Question : 26
Total: 35
SECTION - C
(a) Two charged conducting spheres of radii
OR
(b) A parallel plate capacitor (A) of capacitance
(i) final charges on
(ii) total electrostatic energy stored in A and B finally and that stored in A initially.
Solution: 👈: Video Solution
(a) For sphere with radius a, Potential = V
(since, both the charged spheres are connected by a wire)
Charge = Q a
Capacitance = C a
Electric field, = E a =
For sphere with radiusb
Potential= V
(since both the charged spheres are connected by a wire)
Charge = Q b
Capacitance = C b
Electric field = E b
=
Now,
=
×
=
Or,
=
Or,
=
(since,C d ∕ C b = a ∕ b
Putting in equation (1)
= (
) × (
)
∴
=
OR
(b) (i) Initially,
Charge on A capacitor= CV
Charge onB = 0
After connecting A with B,
Final potential = V ′ =
=
=
Final charge on A capacitor after redistribution
= C V ′ =
=
Final charge on B capacitor after redistribution
= 2 C V ′ =
=
So, the ratio of charges=
= 1 : 2
(ii) Initially energy stored inA =
CV 2
Finally energy stored inA =
C V 2
=
× C × (
) 2
=
× C × (
) 2
=
Finally energy stored inB =
2 CV 2
=
× 2 C × (
) 2
=
× 2 C × (
) 2
=
Total energy stored in A and B
=
+
=
So, the required ratio=
=
=
(since, both the charged spheres are connected by a wire)
For sphere with radius
Potential
(since both the charged spheres are connected by a wire)
(since,
Putting in equation (1)
OR
(b) (i) Initially,
Charge on A capacitor
Charge on
After connecting A with B,
Final charge on A capacitor after redistribution
Final charge on B capacitor after redistribution
So, the ratio of charges
(ii) Initially energy stored in
Finally energy stored in
Finally energy stored in
Total energy stored in A and B
So, the required ratio
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