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$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \;cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7}\; C?$ The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Solution
$(a)$ Charge on sphere $A , q _{ A }=6.5 \times 10^{-7}\, C$
Charge on sphere $B , q _{ B }=6.5 \times 10^{-7} \,C$
Distance between the spheres, $r=50 \,cm =0.5 \,m$ Force of repulsion between the two spheres
$F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}$
Where, $\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,Nm ^{2} \,C ^{-2}$
Therefore,
$F =\frac{9 \times 10^{9} \times\left(6.5 \times 10^{-7}\right)^{2}}{(0.5)^{2}}$
$=1.52 \times 10^{-2} \,N$
Therefore, the force between the two spheres is $1.52 \times 10^{-2} \,N$
$(b)$ After doubling the charge, Charge on sphere $A , q _{ A }=1.3 \times 10^{-6} \,C$
Charge on sphere $B , q _{ B }=1.3 \times 10^{-6} \,C$
The distance between the spheres is halved.
$\therefore r=\frac{0.5}{2}=0.25\, m$
Force of repulsion between the two spheres,
$F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}$$=\frac{9 \times 10^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^{2}}$
$=16 \times 1.52 \times 10^{-2}$
$=0.243 \,N$
Therefore, the force between the two spheres is $0.243 \,N$.
