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(D) The electric field intensity $E$ due to a point charge $Q$ at a distance $R$ is given by $E = \frac{k|Q|}{R^2} = 500 \, Vm^{-1}$.The electric pote

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

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(D) The electric field intensity $E$ due to a point charge $Q$ at a distance $R$ is given by $E = \frac{k|Q|}{R^2} = 500 \, Vm^{-1}$.The electric potential $V$ due to a point charge $Q$ at a distance $R$ is given by $V = \frac{kQ}{R} = -3000 \, V$.Taking the magnitude of potential,$|V| = \frac{k|Q|}{R} = 3000 \, V$.Dividing the expression for $E$ by the expression for $|V|$:$\frac{E}{|V|} = \frac{k|Q|/R^2}{k|Q|/R} = \frac{1}{R} = \frac{500}{3000} = \frac{1}{6}$.Therefore,$R = 6 \, m$.Substituting $R = 6 \, m$ into the potential equation:$|V| = \frac{k|Q|}{R} \Rightarrow 3000 = \frac{9 	imes 10^9 	imes |Q|}{6}$.$|Q| = \frac{3000 	imes 6}{9 	imes 10^9} = \frac{18000}{9 	imes 10^9} = 2 	imes 10^{-6} \, C = 2 \, \mu C$.Thus,the distance is $6 \, m$ and the magnitude of the charge is $2 \, \mu C$.

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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