$(a)$ 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?
$(a)$ 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 $0.243 \,N$.
A particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then time period will be given as
(Consider $k$ as Coulomb's constant)
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How did Coulomb find the law of value of electric force between two point charges ?
As shown in the figure. a configuration of two equal point charges $\left( q _0=+2 \mu C \right)$ is placed on an inclined plane. Mass of each point charge is $20\,g$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $h = x \times 10^{-3}\,m$ The value of $x$ is $..........$.(Take $\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\,Nm ^2 C ^{-2}, g=10\,ms ^{-1}\right)$
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