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Q.
The separation between the two charges $ +q $ and $ -q $ becomes double. The value of force will be:
Rajasthan PMTRajasthan PMT 2004Electric Charges and Fields
Solution:
The force between two charges $q$ and $-q$ is
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \times-q}{r^{2}}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$
where $r=$ separation between the charges
$\therefore F'=\frac{1}{4 \pi \varepsilon_{0}} \frac{q \times-q}{r'^{2}}$
$=-\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q}{(2 r)^{2}} $
$\left(\because r'=2 r\right) $
$=\frac{1}{4} \,F$