If $2 \times 10^{-3} J$ of work is done in moving a particle carrying a charge of $10 \times 10^{-6}$ coulomb from infinity to a point $P .$ What will be the potential at the point $P ?$

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We know that $V=\frac{W}{q}=\frac{2 \times 10^{-3}}{10 \times 10^{-6}}=200 V$.

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