A) -4Q/3
B) -8Q/11
C) -5Q/3
D) -3Q/7
Correct Answer: B
Solution :
[b] From given conditions, \[{{V}_{A}}={{V}_{B}}\]and \[{{V}_{B}}=0\] \[\Rightarrow {{V}_{B}}=\frac{k\left( Q-{{q}_{1}} \right)}{3\alpha }+\frac{k{{q}_{2}}}{2\alpha }+\frac{k{{q}_{1}}}{2\alpha }=0\] \[\Rightarrow 2Q+{{q}_{1}}+3{{q}_{2}}=0\] ?.(i) Using \[{{V}_{A}}={{V}_{C}}\] \[\frac{k\left( Q-{{q}_{1}} \right)}{3\alpha }+\frac{k{{q}_{2}}}{2\alpha }+\frac{k{{q}_{1}}}{2\alpha }=\frac{k{{q}_{1}}}{2\alpha }+\frac{k\left( Q-{{q}_{1}} \right)}{3\alpha }+\frac{k{{q}_{2}}}{2\alpha }\] \[\Rightarrow {{q}_{1}}=-\frac{{{q}_{2}}}{4}\] ?(ii) Using it in (1), \[{{q}_{2}}=-\frac{8}{11}Q\]You need to login to perform this action.
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