Derive an expression for capacitance of a parallel plate capacitor when a conducting slab is inserted between the plates of a capacitor.

To derive the expression for the capacitance of a parallel plate capacitor when a conducting slab is inserted between the plates, we start with the basic principles of capacitance.

Capacitance Basics

The capacitance \( C \) of a parallel plate capacitor is given by the formula:

C = \frac{Q}{V}

where \( Q \) is the charge on the plates and \( V \) is the voltage across them.

Effect of the Conducting Slab

When a conducting slab of thickness \( d \) is inserted between the plates, it effectively divides the capacitor into two smaller capacitors in series. The distance between the plates is \( D \), and the distance between the slab and each plate becomes \( \frac{D - d}{2} \).

Capacitance of Each Section

• The capacitance of the upper section (from the top plate to the slab) is:

C_1 = \frac{\varepsilon_0 A}{\frac{D - d}{2}}

• The capacitance of the lower section (from the slab to the bottom plate) is:

C_2 = \frac{\varepsilon_0 A}{\frac{D - d}{2}}

Combined Capacitance

Since these two capacitances are in series, the total capacitance \( C \) can be calculated using the formula for capacitors in series:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

Substituting the values of \( C_1 \) and \( C_2 \):

\frac{1}{C} = \frac{1}{\frac{\varepsilon_0 A}{\frac{D - d}{2}}} + \frac{1}{\frac{\varepsilon_0 A}{\frac{D - d}{2}}}

This simplifies to:

\frac{1}{C} = \frac{2}{\frac{\varepsilon_0 A}{\frac{D - d}{2}}}

Thus, the total capacitance becomes:

C = \frac{\varepsilon_0 A}{D - d}

Final Expression

In conclusion, the capacitance of a parallel plate capacitor with a conducting slab inserted is:

C = \frac{\varepsilon_0 A}{D - d}

This expression shows how the presence of the conducting slab alters the effective distance between the plates, thereby affecting the capacitance.