A) \[{{Q}^{2}}/(4\pi {{\varepsilon }_{0}}{{a}^{2}})\]
B) \[-{{Q}^{2}}/(4\pi {{\varepsilon }_{0}}{{a}^{2}})\]
C) Zero
D) \[{{Q}^{2}}/(2\pi {{\varepsilon }_{0}}{{a}^{2}})\]
Correct Answer: C
Solution :
\[\left| \overrightarrow{{{F}_{B}}} \right|=\left| \overrightarrow{{{F}_{C}}} \right|=k.\,\frac{{{Q}^{2}}}{{{a}^{2}}}\] Hence force experienced by the charge at A in the direction normal to BC is zero.You need to login to perform this action.
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