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Electrostatics — Physics STD 12 — Question

Tamilnadu BoardEnglish MediumSTD 12PhysicsElectrostatics5 Marks
Question
Two particles having charges Q1 and Q2 when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
✓

Answer

$
F =\frac{1}{4 \pi \varepsilon_0} \frac{ Q _1 Q _2}{r^2}
$
If the distance is educed by half and two particles of charges are doubled.
$\begin{aligned} & F ^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(2 Q _1\right)\left(2 Q _2\right)}{(r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{4 Q _1 Q _2}{\left(r^2 / 4\right)^2} \\ & F ^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{16\left( Q _1 Q _2\right)}{r^2}=16\left[\frac{1}{4 \pi \varepsilon_0} \frac{ Q _1 Q _2}{r^2}\right] \\ & F ^{\prime}=16 F \end{aligned}$

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