JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 5)
Charges –q and +q located at A and B, respectively, constitude 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' = $$\left( {{y \over 3}} \right)$$, the force on Q will be close to - $$\left( {{y \over 3} > > 2a} \right)$$
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9F
3F
F/3
27F
Explanation
Electric field of equitorial plane of dipole
$$ = - {{K\overrightarrow P } \over {{r^3}}}$$
$$ \therefore $$ At P, F $$ = - {{K\overrightarrow P } \over {{r^3}}}$$Q.
At P1 , F1 $$ = - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$$
$$ = - {{K\overrightarrow P } \over {{r^3}}}$$
$$ \therefore $$ At P, F $$ = - {{K\overrightarrow P } \over {{r^3}}}$$Q.
At P1 , F1 $$ = - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$$
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