question_answer

Two square metallic plates of side a = 1 m are kept at d = 8.85 mm apart, like a parallel plate capacitor, in such a way that their surfaces are normal to the oil surface in a tank filled with insulating oil (K = 11). The plates are connected to a battery of emf V = 500 volt as shown in figure. The plates are then lowered vertically into the oil at a speed of\[\text{V}={{10}^{-3}}m{{s}^{-1}}\]. Neglecting resistance of connecting wires, calculate the current drawn from battery during the process (so\[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}m{{~}^{-2}}\])

A) \[5\times {{10}^{-19}}A\]      

B)        \[0.5\times {{10}^{-19}}A\]

C) \[5\times {{10}^{-9}}A\]       

D)        None of these

Correct Answer: C

Solution :

[c] \[C={{C}_{1}}+{{C}_{2}}\] \[=\frac{{{\in }_{0}}a(a-x)}{d}+\frac{{{\in }_{0}}kax}{d}\] \[q=CV\] \[i=\frac{dq}{dt}=V\frac{dC}{dt}\]