Q83 / 120 · VITEEE 2006 · Physics
Figure shows a triangular array of three point charges. The electric potential $\mathrm{V}$ of these source charges at the midpoint $\mathrm{P}$ of the base of the triangle is
$\left[\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right]$
- A$55 \mathrm{kV}$
- Correct answerB$45 \mathrm{kV}$
- C$63 \mathrm{kV}$
- D$49 \mathrm{kV}$
Solution
The net electric potential is algebraic sum of potential due to individual point charges.
$\begin{aligned}
\mathrm{V} &=\frac{1}{4 \pi \in_{0}}\left[\frac{\mathrm{q}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{q}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{q}_{3}}{\mathrm{r}_{3}}\right] \\
\mathrm{V} &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{1 \times 10^{-6}}{0.2}-\frac{2 \times 10^{-6}}{0.2}+\frac{3 \times 10^{-6}}{0.3}\right] \\
&=9 \times 10^{9}[5-10+10] \times 10^{-6} \\
&=9 \times 10^{3}[5]=45 \times 10^{3} \mathrm{~V}=45 \mathrm{kV}
\end{aligned}$
Hence, the correct answer is option: 'B'.