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2. Electric Potential and Capacitance
easy
The potentials of the two plates of capacitor are $+10\,V$ and $-10\, V$. The charge on one of the plates is $40 \,C$. The capacitance of the capacitor is........$F$
A
$2$
B
$4$
C
$0.5$
D
$0.25$
Solution
(a) The potential difference across the parallel plate capacitor is $10\,V – ( – 10\,V) = 20\,V.$
Capacitance $ = \frac{Q}{V} = \frac{{40}}{{20}} = 2\,F.$
Standard 12
Physics
