A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates as shown in the figure. The total capacitance of the capacitor is

Question

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers $1/3$ of the area of its plates as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ . The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option.

TestHub · Accepted Answer

The given arrangement can be considered as two capacitors connected in parallel.

Let the total area of the plates be $A$ and the distance between them be $d$ .

The capacitance of the portion with the dielectric (covering $1/3$ of the area) is:
$C_{1} = \frac{K ϵ _{0} ( A /3 )}{d} = \frac{K ϵ _{0} A}{3 d}$

The capacitance of the remaining portion (covering $2/3$ of the area) is:
$C_{2} = \frac{ϵ _{0} ( 2 A /3 )}{d} = \frac{2 ϵ _{0} A}{3 d}$

The total capacitance $C$ is the sum of $C_{1}$ and $C_{2}$ since they are in parallel:
$C = C_{1} + C_{2} = \frac{K ϵ _{0} A}{3 d} + \frac{2 ϵ _{0} A}{3 d} = \frac{ϵ _{0} A}{3 d} (K + 2)$

Now, finding the ratio $\frac{C}{C _{1}}$ :
$\frac{C}{C _{1}} = \frac{\frac{ϵ _{0} A}{3 d} ( K + 2 )}{\frac{K ϵ _{0} A}{3 d}} = \frac{K + 2}{K}$

Checking other parameters:
Since the two portions are in parallel, the potential difference $V$ across them is the same. The electric field in both regions is $E = \frac{V}{d}$ . Therefore, $E_{1} = E_{2}$ , which gives $\frac{E _{1}}{E _{2}} = 1$ .

The charges on the two portions are $Q_{1} = C_{1} V$ and $Q_{2} = C_{2} V$ .
The ratio of charges is $\frac{Q _{1}}{Q _{2}} = \frac{C _{1}}{C _{2}} = \frac{\frac{K ϵ _{0} A}{3 d}}{\frac{2 ϵ _{0} A}{3 d}} = \frac{K}{2}$ .

Answer: $\frac{C}{C _{1}} = \frac{2 + K}{K}$

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