Two identical condncting spheres, fixed in space, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. A thin conducting wire then connects the spheres, When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the initial charges on the spheres?

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Solution

initial charges = $q_{1} a n d - q_{2}$
Force of attraction = $0.108 N = \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{{0.5}^{2}}$
charges after the connection is made and removed = $\frac{(q_{1} - q_{2})}{2} a n d \frac{(q_{1} - q_{2})}{2}$
Force of repulsion = $0.036 N = \frac{1}{4 π ϵ} \frac{(q_{1} - q_{2})^{2}}{4 \times {0.5}^{2}}$
therefore,
$q_{1} q_{2} = 3.0012 \times 10^{- 12}$ and $(q_{1} - q_{2})^{2} = 4 \times 10^{- 12}$
$\Rightarrow q_{1} = q_{2} = 1 μ F$

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initial charges = q1and−q2Force of attraction = 0.108N=14πϵq1q20.52charges after the connection is made and removed = (q1−q2)2and(q1−q2)2 Force of repulsion = 0.036N=14πϵ(q1−q2)24×0.52therefore, q1q2=3.0012×10−12 and (q1−q2)2=4×10−12⇒q1=q2=1μF