Electrostatics Ques 141
A parallel plate capacitor has $1 \mu \mathrm{F}$ capacitance. One of its two plates is given $+2 \mu \mathrm{C}$ charge and the other plate $-2 \mu \mathrm{C}$ charge. The potential difference developed across the capacitor is
(Main 2019, 8 April II)
(a) $1 \mathrm{~V}$
(b) $5 \mathrm{~V}$
(c) $2 \mathrm{~V}$
(d) $3 \mathrm{~V}$
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Answer:
Correct Answer: 141.(a)
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
Net value of charge on plates of capacitor after steady state is reached is zero
$ q_{\mathrm{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, $q_{2}=4 \mu \mathrm{C}, q_{1}=2 \mu \mathrm{C}$
$\therefore$ Charge of capacitor is $q=\Delta q_{\text {net }}=\frac{4-2}{2}=1 \mu \mathrm{C}$
Potential difference between capacitor plates is
$ V=\frac{Q}{C}=\frac{1 \mu \mathrm{C}}{1 \mu \mathrm{F}}=1 \mathrm{~V} $
BETA
