Charges -q and +q located at A and B. respectively, constitute an electric dipole. Distance \[AB=2a\], 0 is the mid-point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where \[OP=y\] and \[y\,\,>\,\,>\,\,2a\]. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P such that \[OP'=\left( \frac{y}{3} \right)\], the force on Q will be close to- \[\left( \frac{y}{3}>>2a \right)\] [JEE Main 10-Jan-2019 Evening] |
A) 9F
B) 3F
C) F/3
D) 27F
Correct Answer: A
Solution :
[a] |
\[\,\left| \overrightarrow{P} \right|=q.2a\] |
\[F=QE\] |
\[F\,\,=\,\,Q.\frac{2Kp}{{{y}^{3}}}\] |
\[\frac{F}{F'}=\,\frac{Q.\frac{2Kp}{{{y}^{3}}}}{Q.\frac{2Kp}{{{(y/3)}^{3}}}}\] |
\[\frac{F}{F'}=\,\frac{1}{27}\] |
\[F'=27\,F\] |
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