Question

A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance C. If the sheet partly occupies the space between the plates, the capacitance :

• A. Decreases
• B. Increases
• C. Remains the same
• D. Becomes zero

Answer 1

Answer: (B)

Increases

Answer 2

Let the original capacitance of the parallel plate capacitor be $C = \frac{\varepsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is the separation between them.

When a metal sheet of thickness $t$ is inserted between the plates, parallel to them, it effectively reduces the distance over which the electric field can exist. The electric field lines terminate on the surface of the metal sheet and originate from its opposite surface. Therefore, the region occupied by the metal sheet does not contribute to the capacitance in the same way as a dielectric. The effective separation between the plates becomes $d' = d - t$.

The new capacitance $C'$ is given by: $C' = \frac{\varepsilon_0 A}{d'} = \frac{\varepsilon_0 A}{d - t}$

The problem states that the sheet "partly occupies the space between the plates", which implies that the thickness of the metal sheet $t$ is greater than zero ($t > 0$) and less than the plate separation ($t < d$).

Since $t > 0$, the denominator $d-t$ is less than $d$: $d - t < d$

This leads to: $\frac{1}{d - t} > \frac{1}{d}$

Multiplying by $\varepsilon_0 A$ (which is positive), we get: $\frac{\varepsilon_0 A}{d - t} > \frac{\varepsilon_0 A}{d}$

Thus, $C' > C$. The capacitance of the parallel plate capacitor increases when a metal sheet is inserted between its plates.