Calculate the change in potential energy of a particle of charge $+ q$ that is brought from a distance of $3 r$ to a distance of $2 r$ in the electric field of charge $- q$ ?

k q^{2} / r

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- k q^{2} / 6 r

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k q^{2} / 4 r^{2}

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- k q^{2} / 4 r^{2}

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k q^{2} / r^{2}

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Solution

The correct option is C $- k q^{2} / 6 r$
Given : $P B = 2 r$ $P A = 3 r$
Potential at point A $V_{A} = \frac{k q}{P A} = \frac{k q}{3 r}$

Potential at point B $V_{B} = \frac{k q}{P B} = \frac{k q}{2 r}$
Change in potential $Δ V = V_{B} - V_{A} = \frac{k q}{2 r} - \frac{k q}{3 r} = \frac{k q}{6 r}$
Thus change in potential energy by moving a charge $- q$ from A to B $W = - q (Δ V) = \frac{- k q^{2}}{6 r}$

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Calculate the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r in the electric field of charge −q?

The correct option is C −kq2/6rGiven : PB=2r PA=3rPotential at point A VA=kqPA=kq3rPotential at point B VB=kqPB=kq2rChange in potential ΔV=VB−VA=kq2r−kq3r=kq6rThus change in potential energy by moving a charge −q from A to B W=−q(ΔV)=−kq26r

Calculate the change in potential energy of a particle of charge $+ q$ that is brought from a distance of $3 r$ to a distance of $2 r$ in the electric field of charge $- q$ ?