Question

A capacitor is composed of three parallel conducting plates. All three plates are of the same area $A$ . The first pair of plates are kept a distance $d_{1}$ apart, and the space between them is filled with a medium of a dielectric $K_{1}$ . The corresponding data for the second pair are $d_{2}$ and $K_{2}$ , respectively. The middle plate is connected to the positive terminal of a constant voltage source $V$ and the external plates are connected to the other terminal of $V$ .

$(a)$ Find the capacitance of the system.

$(b)$ What is the surface charge density on the middle plate?

$(c)$ Compute the energy density in the medium $K_{1}$ .

Accepted Answer

$(a)$ The equivalent circuit for two capacitors formed by these three plates is as

where the leftmost plate is marked $1$ , middle one as $2$ and rightmost as $3$ .

Let $C_{1}$ be the capacitance of the capacitor formed by plate $1$ , $2$ and $C_{2}$ be the capacitance of the capacitor formed by plate $2.3$ .

Clearly $C_{1} = \frac{K_{1} A \in_{0}}{d_{1}}$

$C_{2} = \frac{K_{2} A \in_{0}}{d_{2}}$ .

Since they are connected to the same voltage.

$C_{eq} = C_{1} + C_{2} = \frac{K_{1} A \in_{0}}{d_{1}} + \frac{K_{2} A \in_{0}}{d_{2}} = A \in_{0} [\frac{K_{1}}{d_{1}} + \frac{K_{2}}{d_{2}}]$

In fact, this is the capacitance of the system.

$(b)$ Now surface charge density is defined as charge per unit area.

Let $Q_{1}$ be the total charge for the first capacitor and $Q_{2}$ be the total charge associated with the second capacitor, then

$Q_{1} = C_{1} V = \frac{K_{1} A \in_{0}}{d_{1}} V$

$Q_{1} = C_{2} V = \frac{K_{2} A \in_{0}}{d_{2}} V$ .

Since the plate $2$ is common in both the capacitors, therefore, the total charge on the plate $2$ is given by

$Q = Q_{1} + Q_{2} = C_{1} V + C_{2} V = \frac{K_{1} A \in_{0}}{d_{1}} V + \frac{K_{2} A \in_{0}}{d_{2}} V = A \in_{0} V [\frac{K_{1}}{d_{1}} + \frac{K_{2}}{d_{2}}]$ .

Hence, the charge density on the plate $2$ $= \frac{Q}{A} = \frac{A V \in_{0} [\frac{K_{1}}{d_{1}} + \frac{K_{2}}{d_{2}}]}{A} = V \in_{0} [\frac{K_{1}}{d_{1}} + \frac{K_{2}}{d_{2}}]$ .

$(c)$ Now energy density in the medium $K_{1}$ is given by

Energy density $= \frac{\frac{1}{2} C_{1} V^{2}}{Volume} = \frac{\frac{1}{2} A \in_{0} \frac{K_{1}}{d_{1}} V^{2}}{A \times d_{1}} = \frac{1}{2} \in_{0} \frac{K_{1}}{{d_{1}}^{2}} V^{2}$ .

Important Questions on Capacitance

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