A) \[\frac{Q({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}{{{4}_{\pi {{\varepsilon }_{0}}}}({{a}^{3}}+{{b}^{3}}+{{c}^{3}})}\]
B) \[\frac{Q}{{{4}_{\pi {{\varepsilon }_{0}}}}(a+b+c)}\]
C) \[\frac{Q}{{{12}_{\pi \varepsilon {{\,}_{0}}}}}\,\,\frac{ab+bc+ca}{abc}\]
D) \[\frac{Q(a+b+c)}{{{4}_{\pi \varepsilon {{\,}_{0}}}}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\]
Correct Answer: D
Solution :
\[x+y+z=Q\] \[\frac{x}{4\pi {{a}^{2}}}=\frac{y}{4\pi {{b}^{2}}}=\frac{z}{4\pi {{c}^{2}}}\] (surface charge density for all are same) \[x:y:z={{a}^{2}}:{{b}^{2}}:{{c}^{2}}\] \[x=\frac{{{a}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}};\,\,\,y\,=\,\frac{{{b}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}};\] \[z=\frac{{{c}^{2}}Q}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] V at \[r<a\] \[=\,\frac{kx}{a}+\frac{ky}{b}+\frac{kz}{c}\] \[=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{Qa}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\,+\,\frac{Qb}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\,+\,\,\frac{Qc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right]\]\[=\,\,\frac{Q}{4\pi {{\in }_{0}}}\,\,\left[ \frac{a+b+c}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \right]\]You need to login to perform this action.
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