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2. Electric Potential and Capacitance
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At a certain distance from a point charge, the field intensity is $500\, Vm^{-1}$ and the potential is $-3000\, V$. The distance to the charge and the magnitude of the charge respectively are
A$6\,m$ and $6\,\mu C$
B$4\,m$ and $2\,\mu C$
C$6\,m$ and $4\,\mu C$
D$6\,m$ and $2\,\mu C$
Solution
$E=\frac{k Q}{R^{2}}=500$
$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=3000 \Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{6} \Rightarrow \mathrm{R}=6 \mathrm{\,m}$
$\frac{\mathrm{kQ}}{6}=3000 \Rightarrow \mathrm{Q}=\frac{6 \times 3000}{9 \times 10^{9}}=2\, \mu \mathrm{C}$
$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=3000 \Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{6} \Rightarrow \mathrm{R}=6 \mathrm{\,m}$
$\frac{\mathrm{kQ}}{6}=3000 \Rightarrow \mathrm{Q}=\frac{6 \times 3000}{9 \times 10^{9}}=2\, \mu \mathrm{C}$
Standard 12
Physics
