Question

A highly conducting sheet of aluminum foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates at distance \[\dfrac{d}{2}\] from the positive plate where d is the distance between the plates. If the capacitance before the insertion of foil was \[10\mu F\], its value after the insertion of foil will be:
A. \[20\mu F\]
B. \[10\mu F\]
C. \[5\mu F\]
D. Zero

Answer 1

Hint: Use the formula for capacitance of the capacitor in terms of distance of separation between the plates. After inserting the aluminum foil between the plates of the capacitor, the capacitor will behave as two capacitors are connected in series. Find the equivalent capacitance of this series combination.

Formula used:
The capacitance of the capacitor is given as,
\[C = \dfrac{{{\varepsilon _0}A}}{d}\]
Here, \[{\varepsilon _0}\] is the permittivity of the free space, A is the area of the capacitor plate and d is the separation between the parallel plates.

Complete step by step answer:
We have given that the aluminum foil of negligible thickness is inserted midway between the two plates of the capacitor. Therefore, the circuit will act as two capacitors are connected in series.
We know that the capacitance of the parallel plate capacitor is given as,
\[C = \dfrac{{{\varepsilon _0}A}}{d}\]
Here, \[{\varepsilon _0}\] is the permittivity of the free space, A is the area of the capacitor plate and d is the separation between the parallel plates.
We assume the capacitance of the first capacitor is \[{C_1}\] and capacitance of the second capacitor is \[{C_2}\].
Now, we can express the capacitance of the two capacitors for which the distance of separation is \[\dfrac{d}{2}\].
\[{C_1} = {C_2} = \dfrac{{{\varepsilon _0}A}}{{\left( {\dfrac{d}{2}} \right)}}\] …… (1)

We have given, the initial capacitance was\[10\mu F\]. Therefore,
 \[C = \dfrac{{{\varepsilon _0}A}}{d} = 10\,\mu F\]
We have the equivalent capacitance of the two capacitors connected in series is,
\[\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}\]
 Therefore, from equation (1) we can write,
\[\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{\dfrac{{{\varepsilon _0}A}}{{\left( {\dfrac{d}{2}} \right)}}}} + \dfrac{1}{{\dfrac{{{\varepsilon _0}A}}{{\left( {\dfrac{d}{2}} \right)}}}}\]
\[ \Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{d}{{{\varepsilon _0}A}}\]
\[ \therefore{C_{eq}} = \dfrac{{{\varepsilon _0}A}}{d} = 10\,\mu F\]

Therefore, the total capacitance of the plates is not affected by the insertion of aluminum foil.

So, the correct answer is “Option B”.

Note:
When we insert the foil of negligible thickness between the plates of the parallel plate capacitor, the area of plates does not change. In this question, we don’t know in which medium the capacitor plates are placed. We have assumed that the plates are in vacuum and considered the permittivity of the free space.