Electrostatics 5 Question 6
7. A parallel plate capacitor has $1 \mu F$ capacitance. One of its two plates is given $+2 \mu C$ charge and the other plate $+4 \mu C$ charge. The potential difference developed across the capacitor is
(Main 2019, 8 April II)
(a) $1 V$
(b) $5 V$
(c) $2 V$
(d) $3 V$
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Answer:
Correct Answer: 7. 2
Solution:
- Net value of charge on plates of capacitor after steady state is reached is
$$ q _{net}=\frac{q _2-q _1}{2} $$
where, $q _2$ and $q _1$ are the charges given to plates.
(Note that this formula is valid for any polarity of charge.)
Here, $q _2=4 \mu C, q _1=2 \mu C$
$\therefore$ Charge of capacitor is $q=\Delta q _{\text {net }}=\frac{4-2}{2}=1 \mu C$
Potential difference between capacitor plates is
$$ V=\frac{Q}{C}=\frac{1 \mu C}{1 \mu F}=1 V $$
