Question

Two square metallic plates of $1$m side are kept $0.01$ m apart, like a parallel plate capacitor, in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with insulating oil. The plates are connected to a battery of emf $500$V. The plates are then lowered vertically into the oil at a speed of $0.001$ m/s. Calculate the current drawn from the battery during the process. (dielectric constant of oil $=11,{\epsilon }_o=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$)

• A. $4.425\times {10}^{-8}A$.
• B. $4.425\times {10}^{-9}A$.
• C. $4.425\times {10}^{-7}A$.
• D. $4.425\times {10}^{-5}A$.

Answer 1

Answer: (B)

$4.425\times {10}^{-9}A$.

$\text{Given}:$ d = 0.01m, V = 500 V, v = 0.001 m/s

$\text{Solution}:$

We know,

$C=\frac{{\epsilon }_0(1-x.1)}{d}+\frac{K{\epsilon }_{0}x}{d}$

$C=\frac{\epsilon 0}{d}(1-x+Kx)$

$C=\frac{{\epsilon }_0}{d}(1+(K-1\left)x\right)$

$\frac{dC}{dt}=\frac{{\epsilon }_0}{d}(K-1)v=\frac{8.85\times {10}^{-12}}{0.01}\times (11-1)\times 0.001=8.85\times {10}^{-12}....\left(1\right)$

Q = CV

$I=V.\frac{dC}{dt}$ ...(2)

From (1) and (2)

$I=500\times 8.85\times {10}^{-2}=44.25\times {10}^{-10}=4.425\times {10}^{-9}Amp$

$\text{Hence B is the correct option}$