A) +Q/4
B) +Q/2
C) -Q/2
D) -Q/4
Correct Answer: D
Solution :
If the system containing three charges in equilibrium then, \[\frac{Qq}{{{r}^{2}}}=\frac{QQ}{{{(2r)}^{2}}}=0\] \[\frac{Qq}{{{r}^{2}}}=-\frac{{{Q}^{2}}}{4{{r}^{2}}}\] \[q=-\frac{Q}{4}\]You need to login to perform this action.
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