Two plates are 2,cm apart, a potential difference of 10volt is applied between them, electric field

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2. Electric Potential and Capacitance

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Two plates are $2\,cm$ apart, a potential difference of $10\;volt$ is applied between them, the electric field between the plates is.........$N/C$

A

$20$

B

$500$

C

$5$

D

$250$

Solution

(b) $E = \frac{V}{d} = \frac{{10}}{{2 \times {{10}^{ – 2}}}} = 500\,N/C$

Standard 12

Physics

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