Question

A parallel plate isolated condenser consists of two metal plates of area

A

and separation

d

. A slab of thickness

t

and dielectric constant

K

is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capacitance of the system. If

K = 2,

for what value of

t / d

will the capacitance of the system be

3 / 2

times that of the condenser with air filling the full space? Calculate the ratio of the energy in the two cases and account for the energy change (assuming

q

charge on the plate to be constant).

Accepted Answer

Let us assume a parallel plate capacitor consisting of plates of area $A$ and separation $d$ with a slab of thickness $t$ inserted between the plates as show in figure.

This is equivalent to two capacitors connected in series with width of one capacitor as $t$ filled with dielectric of dielectric constant $K$ and width of another capacitor as $d - t$ filled with air. Let $C_{1}$ and $C_{2}$ be their capacitances, respectively.

Then, $C_{1} = \frac{K A \in_{\circ}}{t}, C_{2} = \frac{A \in_{\circ}}{d - t}$

Now effective capacitance is given by,

$C_{eff} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{\frac{K A \in_{\circ}}{t} \times \frac{K A \in_{\circ}}{t - d}}{\frac{K A \in_{\circ}}{t} + \frac{K A \in_{\circ}}{t - d}} = \frac{A \in_{\circ}}{d - t (1 - \frac{1}{K})} \dots (i)$

Now according to the question if, i put $K = 2$ in equation $(i)$ , then $C_{eff} = \frac{3}{2} C_{\circ}$ , where $C_{\circ}$ is the capacitance of condenser with air filling full space and we know $C_{\circ} = \frac{A \in_{\circ}}{d}$ .

So we have,

$\frac{2 A \in_{\circ}}{2 d - t} = \frac{3}{2} \frac{A \in_{\circ}}{d} \Rightarrow 4 d = 6 d - 3 t \Rightarrow 2 d = 3 t \Rightarrow \frac{t}{d} = \frac{2}{3} .$

This is the required value of $\frac{t}{d}$ .

Also the ratios of energies in two cases is give by $\frac{E_{o}}{E_{eff}} = \frac{C_{eff}}{C_{\circ}} = \frac{3}{2}$ , because charge is conserved in both a cases, so energy of capacitor is given by $\frac{q^{2}}{2 C}$ . Clearly after inserting, dielectric energy decreases.

Important Questions on Capacitance

Learn and Prepare for any exam you want !