A) \[2\times {{10}^{-7}}N\]
B) Zero
C) 30 N
D) \[2\times {{10}^{-3}}N\]
Correct Answer: B
Solution :
New force\[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(6-2)(2-2)}{{{r}^{2}}}\times {{10}^{-12}}\] \[=0\]You need to login to perform this action.
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