A) \[\frac{Q}{12\pi {{\varepsilon }_{0}}}\frac{ab+bc+ca}{abc}\]
B) \[\frac{Q({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{4\pi {{\varepsilon }_{0}}({{a}^{3}}+{{b}^{3}}+{{c}^{3}})}\]
C) \[\frac{Q}{4\pi {{\varepsilon }_{0}}(a+b+c)}\]
D) \[\frac{Q(a+b+c)}{4\pi {{\varepsilon }_{0}}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\]
Correct Answer: D
Solution :
| \[{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}=Q....(1)\] |
| \[\frac{{{Q}_{1}}}{4\pi {{a}^{2}}}=\frac{{{Q}_{1}}}{4\pi {{b}^{2}}}=\frac{{{Q}_{3}}}{4\pi {{c}^{2}}}=k....(2)\] |
| Subs.\[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}}\]in\[(1)\] |
| \[k=\frac{Q}{4\pi ({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\] |
| \[v=\frac{k{{Q}_{1}}}{a}+\frac{k{{Q}_{2}}}{b}+\frac{k{{Q}_{3}}}{c}.\] |
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