A) A
B) B
C) C
D) D
Correct Answer: B
Solution :
\[{{E}_{axis}}=\frac{kxQ}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{3/2}}}\] |
\[{{E}_{\max }}\Rightarrow when\frac{dE}{dx}=0\Rightarrow x=\pm \frac{R}{\sqrt{2}}\] |
From \[\frac{dV}{dr}=-E\],- we have |
\[{{V}_{\max }}=\frac{Q}{2\sqrt{6}\pi {{\in }_{0}}}\] |
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