A Charge of 3C experiences a force of 3000N when it is moved in a uniform electric field. What is the potential difference between two points separated by a distance of 1cm in the direction of field

10V

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1000V

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3000V

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9000V

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Solution

The correct option is A 10V
$F = E q = \frac{V}{d} q F = 3000 = \frac{V}{10^{- 2}} 3 V = 10 V$

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In a uniform electric field a charge of $3 C$ experiences a force of $3000 N$ . The potential difference between two points $1 c m$ apart along the electric lines of force will be: