A parallel plate capacitor has a dielectric slab of dielectric constant between its plates that covers of the area of its plates, as shown in the figure. The total capacitance of the capacitor is , while that of the portion with dielectric in between is

Question

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $\frac{1}{3}$ 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/options, ignoring edge effects. [2014]

Smart Achievers · Accepted Answer

(1, 4)

The given capacitor is equivalent to two capacitors in parallel with capacitances

$C_{1} = \frac{K ε_{0} (A / 3)}{d} = \frac{K ε_{0} A}{3 d}$

$C_{2} = \frac{ε_{0} (2 A / 3)}{d} = \frac{2 ε_{0} A}{3 d}$

$A$ = area of each plate and

$d$ = distance between the plates

$∴$ $C = C_{1} + C_{2}$

$= \frac{K ε_{0} A}{3 d} + \frac{2 ε_{0} A}{3 d}$ $= \frac{ε_{0} A}{3 d} (K + 2)$

$∴$ $\frac{C}{C_{1}} = \frac{\frac{ε_{0} A (K + 2)}{3 d}}{\frac{ε_{0} A K}{3 d}} = \frac{K + 2}{K}$

Let $V$ be the potential difference between the plates.

$E_{1} = \frac{V}{d}$ and $E_{2} = \frac{V}{d}$

$∴$ $\frac{E_{1}}{E_{2}} = 1$

$Q_{1} = C_{1} V = \frac{K ε_{0} A}{3 d} V$ and $Q_{2} = C_{2} V = \frac{2 ε_{0} A}{3 d} V$

$∴$ $\frac{Q_{1}}{Q_{2}} = \frac{K}{2}$

Solution

1 $\frac{E_{1}}{E_{2}} = 1$

2 $\frac{E_{1}}{E_{2}} = \frac{1}{K}$

3 $\frac{Q_{1}}{Q_{2}} = \frac{3}{K}$

4 $\frac{C}{C_{1}} = \frac{2 + K}{K}$

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Solution

1 E1E2=1

2 E1E2=1K

3 Q1Q2=3K

4 CC1=2+KK

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1 $\frac{E_{1}}{E_{2}} = 1$

2 $\frac{E_{1}}{E_{2}} = \frac{1}{K}$

3 $\frac{Q_{1}}{Q_{2}} = \frac{3}{K}$

4 $\frac{C}{C_{1}} = \frac{2 + K}{K}$