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Question

$E=\frac{k Q}{R^{2}}=500$

$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=3000 \Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{6} \Rightarrow \mathrm{R}=6 \ma

Vedclass · Accepted Answer

$6\,m$ and $2\,\mu C$

Vedclass · Answer

$E=\frac{k Q}{R^{2}}=500$

$\mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{R}}=3000 \Rightarrow \frac{1}{\mathrm{R}}=\frac{1}{6} \Rightarrow \mathrm{R}=6 \mathrm{\,m}$

$\frac{\mathrm{kQ}}{6}=3000 \Rightarrow \mathrm{Q}=\frac{6 	imes 3000}{9 	imes 10^{9}}=2\, \mu \mathrm{C}$

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

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