Electrostatic force on a small sphere of charge 0.4 μ, C due to another small sphere of charge -0.8

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1. Electric Charges and Fields

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The electrostatic force on a small sphere of charge $0.4 \;\mu\, C$ due to another small sphere of charge $-0.8 \;\mu \,C$ in air is $0.2\; N .$

$(a)$ What is the distance between the two spheres?

$(b)$ What is the force on the second sphere due to the first?

Option A

Option B

Option C

Option D

Solution

$(a)$ Electrostatic force on the first sphere, $F =0.2\, N$ Charge on this sphere, $q_{1}=0.4 \,\mu \,C =0.4 \times 10^{-6}\; C$

Charge on the second sphere, $q_{2}=-0.8 \,\mu \,C =-0.8 \times 10^{-6} \,C$

Electrostatic force between the spheres is given by the relation $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$

Where, $\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,Nm ^{2}\, C ^{-2}$

Therefore, $r^{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{F}$

$=\frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^{9}}{0.2}=144 \times 10^{-4}$

$\Rightarrow r=\sqrt{144 \times 10^{-4}}=12 \times 10^{-2}=0.12\, m$

The distance between the two spheres is $0.12 \,m$

$(b)$ Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is $0.2\, N$.

Standard 12

Physics

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