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Q.

A parallel plate capacitor has $1 μF$ capacitance. One of its two plates is given $+ 2 μC$ charge and the other plate, $+ 4 μC$ charge. The potential difference developed across the capacitor is

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a

$3 V$

b

$2 V$

c

$5 V$

d

$1 V$

answer is D.

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Detailed Solution

Question Image

Potential difference

$V_{1} - V_{2} = (E_{1} - E_{2}) d$

$= (\frac{σ_{1}}{2 ε_{0}} - \frac{σ_{2}}{2 ε_{0}}) d$

$= (\frac{σ_{1}}{2 ε_{0}} - \frac{σ_{2}}{2 ε_{0}}) d = \frac{q_{1} d}{2 {Aε}_{0}} - \frac{q_{2} d}{2 {Aε}_{0}} = \frac{4 - 2}{2 x 1} = 1 V$

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A parallel plate capacitor has 1 μF capacitance. One of its two plates is given +2 μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is