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2. Electric Potential and Capacitance
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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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