Gujarati
2. Electric Potential and Capacitance
easy

A charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. .........$V$ is the potential difference between two points separated by a distance of $1\,cm$

A

$10$

B

$250$

C

$1000$

D

$2500$

Solution

(a) $F = QE = \frac{{QV}}{d}$ $==>$ $5000 = \frac{{5 \times V}}{{{{10}^{ – 2}}}}$ $==>$ $V = 10\,volt$

Standard 12
Physics

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