Question

A metal plate of thickness $$d/2$$ is introduced in between the plates of a parallel plate air capacitor with plate separation of $$d$$. Capacity of the metal plate:
A. Decreases 2 times
B. Increases 2 times
C. Remains same
D. Becomes zero

Answer 1

Hint: Use the formula for capacitance of the parallel plate capacitor. Also use the formula for capacitance of the parallel plate capacitor when a dielectric is introduced between the plates of the capacitor. Use the value of dielectric constant of the metal and substitute it in the formula for capacitance. Hence, determine the new value of the capacitance in terms of the initial capacitance.

Formulae used:
The capacitance $$C$$ of a parallel plate capacitor is given by
$$C={\displaystyle \frac{A{\epsilon }_0}{d}}$$ …… (1)
Here, $$A$$ is an area of the capacitor, $${\epsilon }_0$$ is permittivity of free space and $$d$$ is separation between the two plates of the capacitor.
The capacitance $$C$$ of the parallel plate capacitor when a dielectric is introduced between the plates of the capacitor is
$$C={\displaystyle \frac{A{\epsilon }_0}{d-t+{\displaystyle \frac{t}{K}}}}$$ …… (2)
Here, $$A$$ is the area of capacitor, $${\epsilon }_0$$ is permittivity of free space, $$d$$ is separation between the two plates of the capacitor, $$t$$ is thickness of the dielectric and $$K$$ is dielectric constant.

Complete step by step answer:
We have given that the separation between the two plates of a parallel plate air capacitor is $$d$$.A metal plate of thickness $${\displaystyle \frac{d}{2}}$$ is introduced between the two plates of the capacitor.
$$t={\displaystyle \frac{d}{2}}$$
We have asked to calculate the new capacitance of the capacitor when the metal plate is introduced between the plates of the capacitor. Let $$C$$ be the initial capacitance of the capacitor and $$C^{\prime }$$ be the new capacitance of the capacitor.The initial capacitance of the parallel plate capacitor is given by
$$C={\displaystyle \frac{A{\epsilon }_0}{d}}$$
We know that the dielectric constant for the metal is infinite.
$$K=\mathrm{\infty }$$
Let us now calculate the new capacitance of the capacitor after introduction of the metal plate using equation (2). Substitute $$C^{\prime }$$ for $$C$$, $${\displaystyle \frac{d}{2}}$$ for $$t$$ and $$\mathrm{\infty }$$ for $$K$$ in equation (2).
$$C^{\prime }={\displaystyle \frac{A{\epsilon }_0}{d-{\displaystyle \frac{d}{2}}+{\displaystyle \frac{{\displaystyle \frac{d}{2}}}{\mathrm{\infty }}}}}$$
$$\Rightarrow C^{\prime }={\displaystyle \frac{A{\epsilon }_0}{d-{\displaystyle \frac{d}{2}}+0}}$$
$$\Rightarrow C^{\prime }={\displaystyle \frac{A{\epsilon }_0}{{\displaystyle \frac{d}{2}}}}$$
$$\Rightarrow C^{\prime }={\displaystyle \frac{2A{\epsilon }_0}{d}}$$
Substitute $$C$$ for $${\displaystyle \frac{A{\epsilon }_0}{d}}$$ in the above equation.
$$\therefore C^{\prime }=2C$$
Therefore, the new capacitance of the parallel plate capacitor will increase 2 times.

Hence, the correct option is B.

Note:One can also solve the same question by another method. One can use the concept that when the metal plate of given thickness is introduced between the plates of the capacitor, the charge on one plate of capacitor shifts on the introduced metal plate and the separation between the plates of capacitor decreases to a value decreased by the value equal to thickness of the introduced metal plate. This new separation distance gives the value of new capacitance.