A) \[\frac{Q}{\sqrt{\pi {{\varepsilon }_{0}}RL}}\]
B) \[\frac{Q}{\sqrt{8\pi {{\varepsilon }_{0}}RL}}\]
C) \[\frac{Q}{2\sqrt{\pi {{\varepsilon }_{0}}RL}}\]
D) \[\frac{Q}{\sqrt{2\pi {{\varepsilon }_{0}}RL}}\]
Correct Answer: B
Solution :
[b] At \[i={{i}_{\max }},{{Q}_{1}}-{{Q}_{2}}=\frac{Q}{2}\] \[\therefore \,\,\,\,\,\frac{1}{2}Li_{\max }^{2}=\frac{{{Q}^{2}}}{2C}-2\times \frac{{{(Q/2)}^{2}}}{2C}\] \[\frac{1}{2}Li_{\max }^{2}=\frac{{{Q}^{2}}}{4c}\]You need to login to perform this action.
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