Two particles \({A}\) and \({B}\) having charges \(20\, \mu {C}\) and \(-5\, \mu {C}\) respectively are held fixed with a separation of \(5\, {cm}\). At what position a third charged particle should be placed so that it does not experience a net electric force?

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Answer & Solution

Correct Answer

(B) At \(5\, {cm}\) from \(-5 \,\mu {C}\) on the right side

Step-by-step Solution

Detailed explanation

Null point is possible only right side of \(-5 \,\mu {C}\)

\({E}_{{N}}=+\frac{{k}(-5\, \mu {C})}{{x}^{2}}+\frac{{k}(20 \,\mu {C})}{(5+{x})^{2}}=0\)

\({x}=5\, {cm}\)

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