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Two positively charged particles each having charge `Q` and are `d` distance apart. A third charge is introduced in midway on the line joining the two charges. Find the nature and magnitude of third charge so that the system is in equilibrium. A. `q=(Q)/4`B. `q=Q/4`C. `q=(3Q)/4`D. `q=-(3Q)/4` |
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Answer» (a) Condition for equilibrium `(kQ^(2))/(d^(2))+(kqQ)/((d//2)^(2))=0implies(kQ)/(d^(2))(Q+4q)=0` `q=(-Q)/4` |
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