Question:
Charges -q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O 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' = \(\left(\frac y3\right)\), the force on Q will be close to - \(\left(\frac y3\gg2a\right)\)
Charges -q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O 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' = \(\left(\frac y3\right)\), the force on Q will be close to - \(\left(\frac y3\gg2a\right)\)
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On the equatorial line far from a dipole the field varies as 1/r^3, so the force on Q is proportional to 1/r^3.
Updated On: Nov 5, 2026
- 9F
- 3F
- F/3
- 27F
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The Correct Option is D
Solution and Explanation
For \(r\gg2a\) the equatorial field of a dipole is \(E=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}\), so \(F=QE\propto\frac1{r^3}\).
At P the distance is \(y\) and at \(P'\) it is \(\frac y3\): \(F'=F\left(\frac{y}{y/3}\right)^3=27F\).
Answer: (4) 27F
At P the distance is \(y\) and at \(P'\) it is \(\frac y3\): \(F'=F\left(\frac{y}{y/3}\right)^3=27F\).
Answer: (4) 27F
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