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A charge Q is imparted to two identical capacitors in paralle. Separation of the plates in each capacitor is d_0. Suddenly, the first plate of the first capacitor and the second plate of the second capacitor start moving to the left with speed v, then |
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Answer» charges on the two CAPACITORS as a function of time are `(Q(d_0-VT))/(2d_0), (Q(d_0+vt))/(2d_0)`. Leq `q_(1)` and `q_(2)` be the instantaneous charges on capacitors. SINCE they are in parallel, then `(q_(1))/(C_(1))=(q_(2))/(C_(2))` and `q_(1)+q_(2)=Q` `C_(1)=(epsilon_(0)A)/(d_(0)+vt),C_(2)=(epsilon_(0)A)/(d_(0)-vt)` So `(q_(1))/(q_(2))=(C_(1))/(C_(2))=(d_(0)-vt)/(d_(0)+vt)` or `q_(2)((d_(0)-vt)/(d_(0)+vt))+q_(2)=0` So `q_(2)=(Q(d_(0)+vt))/(2d_(0))` and `q_(1)=(Q(d_(0)-vt))/(2d_(0))` HENCE, option (a) is correct and option (b) is INCORRECT. `i=(-dq_(1))/(dt)` or `(dq_(2))/(dt)` or `i=(Q_(v))/(2d_(0))` Which does not depend on time. So option (d) is correct and option (c) is incorrect. |
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