A) \[\frac{Qv}{2d}\]
B) \[\frac{Qv}{d}\]
C) \[\frac{3Qv}{d}\]
D) \[\frac{2Qv}{d}\]
Correct Answer: B
Solution :
[b] \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d-vt},\,\,{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d+vt}\] \[\because \] Potential difference across two capacitors is always equal. \[\Rightarrow \ \,\,\,\,\frac{q}{{{C}_{1}}}=\frac{(2Q-q)}{{{C}_{2}}}\] \[\Rightarrow \ \,\,\,\,\frac{q(d-vt)}{{{\varepsilon }_{0}}A}=\frac{(2Q-q)\,(d+vt)}{{{\varepsilon }_{0}}A}\] \[\Rightarrow \ \,\,\,\,q=\frac{Q}{d}(d+vt)\] \[\Rightarrow \ \,\,\,\,\frac{dq}{dt}=i=\frac{Qv}{d}\]You need to login to perform this action.
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