Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of $0.108N$ when separated by $50.0cm$, 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 $0.0360N$. 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

Answer: (B)

Given that $F_1=0.108N$
$r=0.5m$
$F_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$0.108=9\times 10^9\frac{q_1{q}_2}{(0.5{)}^2}$
$q_1{q}_2=\frac{0.25\times 0.108}{9\times 10^9}$
$=\frac{108\times 25\times 10^{-5}}{9\times 10^9}$
$q_1{q}_2=3\times 10^{-12}{C}^2\dots$ (i)
and $0.036=9\times 10^9\frac{{\left(\frac{q_1{q}_2}{2}\right)}^2}{(0.5{)}^2}$
$=\frac{0.036\times 4\times 0.25}{9\times 10^9}={\left(q_1-q_2\right)}^2\frac{4\times 4\times 25\times 10^5}{10^9}{\left(q_1-q_2\right)}^2$
$4\times 10^{-12}={\left(q_1-q_2\right)}^{2}2\times 10^{-6}=q_1-q_2\dots$ (iii)
Now ${\left(q_1+q_2\right)}^2={\left(q_1-q_2\right)}^2+4q_1{q}_2$
$=4\times 10^{-12}+4\times 3\times 10^{-12}$
$=16\times 10^{-12}{q}_1+q_2=4\times 10^{-6}\dots$ (iv)
From Eqs. (iii) and (iv), we
get $q_1=-3\times 10^{-6}$ and $q_2=1\times 10^{-6}$