Q133 / 180 · COMEDK 2018 · Physics
Two identieal metal spheres charged with $+12 \mu \mathrm{F}$ and $-8 \mu \mathrm{F}$ are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is
- A$12: 1$
- B$8: 1$
- Correct answerC$24: 1 .$
- D$4: 1$
Solution
$$
\begin{aligned}
Q_{1} &=12 \mu \mathrm{F}=12 \times 10^{-6} \mathrm{~F} \\
Q_{2} &=-8 \mu \mathrm{F} \\
&=-8 \times 10^{-6} \mathrm{~F}
\end{aligned}
$$
Initial value of electrostatic force of attraction between both spheres,
$$
\begin{aligned}
F_{1} &=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1} Q_{2}}{r^{2}} \\
&=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{12 \times 10^{-6} \times 8 \times 10^{-6}}{r^{2}} \\
F_{1} &=\frac{1}{4 \pi \varepsilon_{0}} \frac{96 \times 10^{-12}}{r^{2}}
\end{aligned}
$$
When both charged spheres are kept in contact, then new charges on the spheres will be
$$
\begin{aligned}
Q_{1}^{\prime}=Q_{2}^{\prime} &=\frac{Q_{1}+Q_{2}}{2} \\
&=\frac{12+(-8)}{2}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}
\end{aligned}
$$
$\therefore$ Final value of electrostatic force between the spheres,
$$
\begin{aligned}
F_{2} &=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}^{\prime} \times Q_{2}^{\prime}}{r^{2}} \\
&=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 \times 10^{-6} \times 2 \times 10^{-6}}{r^{2}} \\
F_{2} &=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 \times 10^{-12}}{r^{2}}
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{array}{ll}
\therefore \quad & \frac{F_{1}}{F_{2}}=\frac{\frac{1}{4 \pi \varepsilon_{0}}}{} \cdot \frac{96 \times 10^{-12}}{\frac{1}{4 \pi \varepsilon_{0}}} \cdot \frac{4 \times 10^{-12}}{r^{2}} \\
\Rightarrow \quad & \frac{F_{1}}{F_{2}}=\frac{24}{1} \\
\Rightarrow \quad & F_{1}: F_{2}=24: 1
\end{array}
$$
Hence, the correct answer is option: 'C'.