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  4. Two plates are 2 cm apart. A potential difference of 10 volt is applied between them. The electric field between the plates is

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Q. Two plates are $2\, cm$ apart. A potential difference of $10$ volt is applied between them. The electric field between the plates is

Electrostatic Potential and Capacitance

Solution:

$E=\frac{V}{d}=\frac{10}{2 \times 10^{-2}}$
$=500\, N / C$