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What is the force (in $N$) between two small charged spheres having charges of $2 \times 10^{-7} \;C$ and $3 \times 10^{-7} \;C$ placed $30\; cm$ apart in air?
$3 \times 10^{-4}\; N$
$6 \times 10^{-3}\; N$
$8 \times 10^{-2}\; N$
$1 \times 10^{-3}\; N$
Solution
Repulsive force of magnitude $6 \times 10^{-3}\, N$
Charge on the first sphere, $q_{1}=2 \times 10^{-7} \,C$
Charge on the second sphere, $q_{2}=3 \times 10^{-7} \,C$
Distance between the spheres, $r=30\, cm =0.3 \,m$
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, force
$F =\frac{9 \times 10^{9} \times 2 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} \,N$
Hence, force between the two small charged spheres is $6 \times 10^{-3}\; N$. The charges are of same nature. Hence, force between them will be repulsive.