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2. Electric Potential and Capacitance
easy
The work done in carrying a charge of $5\,\mu \,C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $({V_B} - {V_A})$ is then
A
$+ 2\,kV$
B
$-2\, kV$
C
$+ 200\, V$
D
$-200\, V$
Solution
(a) Work done $W = Q({V_B} – {V_A}) \Rightarrow \,({V_B} – {V_A}) = \frac{W}{Q}$ $ = \frac{{10 \times {{10}^{ – 3}}}}{{5 \times {{10}^{ – 6}}}}J/C = 2\,kV$
Standard 12
Physics
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