Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of $0.108\,N$ when separated by $50.0\, cm$, center to center. The spheres are then connected by a thin conducting wire. When the wire is removed the spheres repel each other with an electrostatic force of $0.0360\, N$. The initial charges on the spheres were

• A. $ 9\times 10^{-6}C,\,-3\times 10^{-6}C $
• B. $ 1\times 10^{-6}C,\,-3\times 10^{-6}C $
• C. $ -3\times 10^{-6}C,\,\,\,2\times 10^{-6}C $
• D. $ 1\times 10^{-6}C,-\,2\times 10^{-6}C $

Answer 1

The Correct Option is B

Given that $F_{1}=0.108 \,N $
$r = 0.5\, m $
$F_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$0.108=9 \times 10^{9} \frac{q_{1} q_{2}}{(0.5)^{2}}$
$ q_{1} q_{2}=\frac{0.25 \times 0.108}{9 \times 10^{9}}$
$=\frac{108 \times 25 \times 10^{-5}}{9 \times 10^{9}}$
$q_{1} q_{2}=3 \times 10^{-12} C^{2} \ldots$ (i)
and $0.036=9 \times 10^{9} \frac{\left(\frac{q_{1} q_{2}}{2}\right)^{2}}{(0.5)^{2}}$
$=\frac{0.036 \times 4 \times 0.25}{9 \times 10^{9}}=\left(q_{1}-q_{2}\right)^{2} \frac{4 \times 4 \times 25 \times 10^{5}}{10^{9}}\left(q_{1}-q_{2}\right)^{2}$
$4 \times 10^{-12}=\left(q_{1}-q_{2}\right)^{2} 2 \times 10^{-6}=q_{1}-q_{2} \ldots$ (iii)
Now $\left(q_{1}+q_{2}\right)^{2}=\left(q_{1}-q_{2}\right)^{2}+4 q_{1} q_{2}$
$=4 \times 10^{-12}+4 \times 3 \times 10^{-12}$
$=16 \times 10^{-12} q_{1}+q_{2}=4 \times 10^{-6} \ldots$ (iv)
From Eqs. (iii) and (iv), we
get $q_{1}=-3 \times 10^{-6}$ and $q_{2}=1 \times 10^{-6}$