JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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?
- A At \(5\, {cm}\) from \(20\, \mu {C}\) on the left side of system
- B At \(5\, {cm}\) from \(-5 \,\mu {C}\) on the right side
- C At \(1.25 \,{cm}\) from \(-5\, \mu {C}\) between two charges
- D At midpoint between two charges
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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