If two similarly charged particles are brought near one another, the potential energy of the system will :

increase

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decrease

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remains the same

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equal to the K.E.

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Solution

The correct option is B increase
Electrostatic $P . E = \frac{K Q_{1} Q_{2}}{r}$ where $K = \frac{1}{4 π ϵ_{o}}$
Now for both similar charges, $P . E > 0$
Thus decreasing r will increase P.E

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If two similarly charged particles are brought near one another, the potential energy of the system will :

The correct option is B increaseElectrostatic P.E=KQ1Q2r where K=14πϵoNow for both similar charges, P.E>0Thus decreasing r will increase P.E

The correct option is B increase
Electrostatic $P . E = \frac{K Q_{1} Q_{2}}{r}$ where $K = \frac{1}{4 π ϵ_{o}}$
Now for both similar charges, $P . E > 0$
Thus decreasing r will increase P.E