Saved Bookmarks
| 1. |
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 A. charges on the two capacitors as a function of time are `(Q(d_0-vt))/(2d_0), (Q(d_0+vt))/(2d_0)`.B. charges on the two capacitors as a function of time are `(Qd_0)/(2(d_0-vt)), (Qd_0)/(2(d_0+vt))`.C. current in the circuit will increase as time passes onD. current in the circuit will be constant. |
|
Answer» Correct Answer - A::D a.,d. 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. |
|