Two identical capacitors are connected as shown and having initial charge Q 0 . Separation between plates of each capacitor is d 0 . Suddenly the left plate of upper capacitor and right plate of lower capacitor start moving with speed v towards left while other plate of each capacitor remains fixed. (given 
= 10 amp). The value of current (in amp) in the circuit is………. × 4 ampere.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(5)
Sol. 
q 1 + q 2 = 2Q 0
C 1 = 
, C 2 = 
⇒ q 2 + q 2 
= 2Q 0
⇒ q 2 
= 2Q 0
⇒ q 2 = 
(d 0 + Vt)
I = 
= 20 amp
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