A parallel plate capacitor is charged up to a potential of 300 volts. Area of the plates is 100 cm 2 and spacing between them is 2 cm. If the plates are moved apart to a distance of 2.5 cm without disconnecting the power source, then ( ∈ 0 = 9 × 10 –12 C 2 N –1 m –2 ):
(i) Electric field inside the capacitor when distance is 2.5 cm:
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) E = 
= 
= 12 × 10 3 V/m
(ii) Δ U = U f – U i = 
C f V 2 – 
C i V 2
= 
V 2
= 
(300) 2
= – 405 × 10 –10 J.
(iii) E = 
= Constant
= 
= 
= 15 × 10 3 V/m.
(iv) Q = 
V = constant
Δ U = 
= 
A ∈ 0 V 2 
= 
V 2 (d f – d i )
= 
= 5.0625 × 10 –8 J
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