Electrostatics 5 Question 48
50. Two square metal plates of side $1 m$ 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 an 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 $0.001 ms^{-1}$. Calculate the current drawn from the battery during the process. (Dielectric constant of oil $=11$,
$$ \left.\varepsilon _0=8.85 \times 10^{-12} C^{2} N^{-1} m^{-2}\right) $$
(1994, 6M)
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Solution:
- Let $a$ be the side of the square plate.
As shown in figure, $C _1$ and $C _2$ are in parallel. Therefore, total capacity of capacitors in the position shown is
$$ \begin{aligned} & C=C _1+C _2 \\ & C=\frac{\varepsilon _0 a(a-x)}{d}+\frac{K \varepsilon _0 a x}{d} \end{aligned} $$
$\therefore \quad q=C V=\frac{\varepsilon _0 a V}{d}(a-x+K x)$
As plates are lowered in the oil, $C$ increases or charge stored will increase.
Therefore, $i=\frac{d q}{d t}=\frac{\varepsilon _0 a V}{d}(K-1) \cdot \frac{d x}{d t}$
Substituting the values
$$ \begin{aligned} \varepsilon _0 & =8.85 \times 10^{-12} C^{2} / N-m^{2} \\ a & =1 m, V=500 \text { volt }, d=0.01 m, K=11 \end{aligned} $$
and $\frac{d x}{d t}=$ speed of plate $=0.001 m / s$
We get current $i=\frac{\left(8.85 \times 10^{-12}\right)(1)(500)(11-1)(0.001)}{(0.01)}$
$$ i=4.43 \times 10^{-9} A $$
