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Physics Question on Electric charges and fields

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N0.108\,N when separated by 50.0 cm50.0\, 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 N0.0360\, N. The initial charges on the spheres were

A

9×10−6C, −3×10−6C9\times 10^{-6}C,\,-3\times 10^{-6}C

B

1×10−6C, −3×10−6C1\times 10^{-6}C,\,-3\times 10^{-6}C

C

−3×10−6C,   2×10−6C-3\times 10^{-6}C,\,\,\,2\times 10^{-6}C

D

1×10−6C,− 2×10−6C1\times 10^{-6}C,-\,2\times 10^{-6}C

Answer

1×10−6C, −3×10−6C1\times 10^{-6}C,\,-3\times 10^{-6}C

Explanation

Solution

Given that F1=0.108 NF_{1}=0.108 \,N
r=0.5 mr = 0.5\, m
F1=14πε0q1q2r2F_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}
0.108=9×109q1q2(0.5)20.108=9 \times 10^{9} \frac{q_{1} q_{2}}{(0.5)^{2}}
q1q2=0.25×0.1089×109q_{1} q_{2}=\frac{0.25 \times 0.108}{9 \times 10^{9}}
=108×25×10−59×109=\frac{108 \times 25 \times 10^{-5}}{9 \times 10^{9}}
q1q2=3×10−12C2…q_{1} q_{2}=3 \times 10^{-12} C^{2} \ldots (i)
and 0.036=9×109(q1q22)2(0.5)20.036=9 \times 10^{9} \frac{\left(\frac{q_{1} q_{2}}{2}\right)^{2}}{(0.5)^{2}}
=0.036×4×0.259×109=(q1−q2)24×4×25×105109(q1−q2)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}
4×10−12=(q1−q2)22×10−6=q1−q2…4 \times 10^{-12}=\left(q_{1}-q_{2}\right)^{2} 2 \times 10^{-6}=q_{1}-q_{2} \ldots (iii)
Now (q1+q2)2=(q1−q2)2+4q1q2\left(q_{1}+q_{2}\right)^{2}=\left(q_{1}-q_{2}\right)^{2}+4 q_{1} q_{2}
=4×10−12+4×3×10−12=4 \times 10^{-12}+4 \times 3 \times 10^{-12}
=16×10−12q1+q2=4×10−6…=16 \times 10^{-12} q_{1}+q_{2}=4 \times 10^{-6} \ldots (iv)
From Eqs. (iii) and (iv), we
get q1=−3×10−6q_{1}=-3 \times 10^{-6} and q2=1×10−6q_{2}=1 \times 10^{-6}