Charges on two plates of a 10,μ f capacitor are 5,μ C and 15,μ C then potential difference across

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2. Electric Potential and Capacitance

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The charges on two plates of a $10\,\mu f$ capacitor are $5\,\mu C$ and $15\,\mu C$ then the potential difference across the capacitor plates is........$V$

A

$0.5$

B

$1$

C

$1.5$

D

$2$

Solution

The charge of the capacitor in expression $Q=C V$ is the magnitude of charge which appears on facing surfaces of plates of capacitor.

The distribution of charge on various faces for present case are shown in figure.

So, here $Q=5\, \mu C$

$\therefore \quad \mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{5 \times 10^{-6}}{10 \times 14^{-6}}=0.5 \mathrm{\,volt}$

Standard 12

Physics

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