A charge 3 coulomb experiences a force 3000 N when placed in a uniform electric field. potential dif

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2. Electric Potential and Capacitance

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A charge $3$ coulomb experiences a force $3000$ $N$ when placed in a uniform electric field. The potential difference between two points separated by a distance of $1$ $cm$ along the field lines is.....$V$

A

$10$

B

$90$

C

$1000 $

D

$9000$

Solution

Electric force $F=q E$

given, $F=3000 N, q=3 C$

$\therefore E=\frac{F}{q}=\frac{3000}{3}=1000 \mathrm{V} / \mathrm{m}$

Potential difference, $V=E d$ where $d=0.01 m$

$\therefore V=1000 \times 0.01=10 V$

Standard 12

Physics

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