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(D) The potential at point $A$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_A = \frac{k(1 	imes 10^{-6})}{2} + \frac{k(-1 	imes 10^{-6})}{3}

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$3000$

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(D) The potential at point $A$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_A = \frac{k(1 	imes 10^{-6})}{2} + \frac{k(-1 	imes 10^{-6})}{3} = k 	imes 10^{-6} \left( \frac{1}{2} - \frac{1}{3} ight) = k 	imes 10^{-6} \left( \frac{1}{6} ight)$$V_A = \frac{9 	imes 10^9 	imes 10^{-6}}{6} = 1.5 	imes 10^3 \ V$The potential at point $B$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_B = \frac{k(1 	imes 10^{-6})}{3} + \frac{k(-1 	imes 10^{-6})}{2} = k 	imes 10^{-6} \left( \frac{1}{3} - \frac{1}{2} ight) = k 	imes 10^{-6} \left( -\frac{1}{6} ight)$$V_B = -\frac{9 	imes 10^9 	imes 10^{-6}}{6} = -1.5 	imes 10^3 \ V$The potential difference between $A$ and $B$ is:$V_A - V_B = 1.5 	imes 10^3 - (-1.5 	imes 10^3) = 3.0 	imes 10^3 \ V = 3000 \ V$

As shown in the figure,positive and negative charges of $1\ \mu C$ are placed at two points. Find the potential difference between $A$ and $B$ in $volt$.

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