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

Question

A parallel plate capacitor has 1 μ F capacitance. One of its two plates is given +2 μ C charge and the other plate, +4 μ C charge. The potential difference developed across the capacitor is:- (A) 5 V (B) 2 V (C) 3 V (D) 1 V

Tardigrade Admin · Accepted Answer

Charges at inner plates are 1 μ C and - 1 μ C ∴ Potential difference across capacitor =(q/c) = (1 μ C/1 μ F) = (1×10-6C/1 ×10-6 Farad) = 1V

Q. A parallel plate capacitor has $1 μ F$ capacitance.
One of its two plates is given $+ 2 μ C$ charge and the other plate, $+ 4 μ C$ charge. The potential difference developed across the capacitor is:-

5 V

2 V

3 V

1 V

Charges at inner plates are $1 μ C$ and $- 1 μ C$
$∴$ Potential difference across capacitor
$= \frac{q}{c} = \frac{1 μ C}{1 μ F} = \frac{1 \times 1 0 ^{- 6} C}{1 \times 1 0 ^{- 6} F a r a d} = 1 V$

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

5 V

2 V

3 V

1 V

Charges at inner plates are 1μC and −1μC ∴ Potential difference across capacitor =cq​=1μF1μC​=1×10−6Farad1×10−6C​=1V

Q. A parallel plate capacitor has $1 μ F$ capacitance.
One of its two plates is given $+ 2 μ C$ charge and the other plate, $+ 4 μ C$ charge. The potential difference developed across the capacitor is:-

Charges at inner plates are $1 μ C$ and $- 1 μ C$
$∴$ Potential difference across capacitor
$= \frac{q}{c} = \frac{1 μ C}{1 μ F} = \frac{1 \times 1 0 ^{- 6} C}{1 \times 1 0 ^{- 6} F a r a d} = 1 V$