Two capacitors of capacitance $2\ \mu F$ and $3\,\mu F$ are joined in series. Outer plate first capacitor is at $1000\, volt$ and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will be......$Volt$
Medium
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(d) Equivalent capacitance $ = \frac{{2 \times 3}}{{2 + 3}} = \frac{6}{5}\,\mu F$
Total charge by $Q = CV$$ = \frac{6}{5} \times 1000 = 1200\,\mu C$
Potential $(V)$ across $2\,\mu F$ is $V = \frac{Q}{C} = \frac{{1200}}{2} = 600\,volt$
Potential on internal plates $ = 1000 - 600 = 400\,V$
Total charge by $Q = CV$$ = \frac{6}{5} \times 1000 = 1200\,\mu C$
Potential $(V)$ across $2\,\mu F$ is $V = \frac{Q}{C} = \frac{{1200}}{2} = 600\,volt$
Potential on internal plates $ = 1000 - 600 = 400\,V$
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