A) 9F
B) 3F
C) F/3
D) 27F
Correct Answer: A
Solution :
\[\,\left| \overrightarrow{P} \right|=q.2a\] \[F=QE\] \[F\,\,=\,\,Q.\frac{2Kp}{{{y}^{3}}}\] \[\frac{F}{F'}=\,\frac{Q.\frac{2Kp}{{{y}^{3}}}}{Q.\frac{2Kp}{{{(y/3)}^{3}}}}\] \[\frac{F}{F'}=\,\frac{1}{27}\] \[F'=27\,F\]You need to login to perform this action.
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