Q169 / 180 · KCET 2022 · Physics
Two tiny spheres carrying charges $1.8 \mu \mathrm{C}$ and $2.8 \mu \mathrm{C}$ arc located at $40 \mathrm{~cm}$ apart. The potential at the mid - point of the line joining the two charges is
- Correct answerA$2.1 \times 10^5 \mathrm{~V}$
- B$1.3 \times 10^4 \mathrm{~V}$
- C$3.6 \times 10^5 \mathrm{~V}$
- D$3.8 \times 10^4 \mathrm{~V}$
Solution
Charge on first sphere, $q_1=1.8 \mu \mathrm{C}=1.8 \times 10^{-6} \mathrm{C}$
Charge on second sphere, $q_2=2.8 \mu \mathrm{C}=2.8 \times 10^{-6} \mathrm{C}$
Distance between the two spheres, $r=40 \mathrm{~cm}$
Distance between mid-point and each sphere $r_1=r_2=20 \mathrm{~cm}=0.2 \mathrm{~m}$
$\therefore$ Electric potential at mid-point due to both charged sphere
$$
\begin{aligned}
V & =V_1+V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2} \\
& =9 \times 10^9 \times \frac{1.8 \times 10^{-6}}{0.2}+9 \times 10^9 \times \frac{2.8 \times 10^{-6}}{0.2} \\
& =9 \times 10^3(9+14)=2.07 \times 10^5 \mathrm{~V} \approx 2.1 \times 10^5 \mathrm{~V}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.