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# When a test charge is placed at the centre of dipole then what will be the net force on the test charge? Please give good explanation.

Ash
36 Points
2 years ago
Let two charges of equal magnitude Q but opposite sign be placed at 2a from each other.
Now a charge q0 is placed midway between them.
The force experienced by it due the negative charge is
$F=\frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ towards the negative charge
The force experienced by it due the positive charge is
$F=\frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ away from the postive charge
So,the net force experienced by it will be
$F_{net}=2 \frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ towards the negative charge.