Ncert Exercises & Solutions

Q 1.2) The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the 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?

(a) Electrostatic force on the first sphere, F = 0.2 N
Charge on this sphere, $q_{1} = 0.4 μ C = 0.4 \times 10^{- 6} C$
Charge on the second sphere, $q_{2} = - 0.8 μ C = - 0.8 \times 10^{- 6} C$
The electrostatic force between the spheres is given by the relation
$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
Where, $ε_{0}$ = Permittivity of free space and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}$
Therefore.

$r^{2} = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{F}$

$\begin{array}{l} = \frac{0.4 \times 10^{- 6} \times 8 \times 10^{- 6} \times 9 \times 10^{9}}{0.2} = 144 \times 10^{- 4} \\ \Rightarrow r = \sqrt{144 \times 10^{- 4}} = 12 \times 10^{- 2} = 0.12 m \end{array}$ 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.

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