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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
$\left[\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right]$
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$\left[\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right]$
Solution:
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Verified Answer
The correct answer is:
$45 \mathrm{kV}$
The net electric potential is algebraic sum of potential due to individual point charges.
$\begin{aligned}
\mathrm{V} &=\frac{1}{4 \pi \in_{0}}\left[\frac{\mathrm{q}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{q}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{q}_{3}}{\mathrm{r}_{3}}\right] \\
\mathrm{V} &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{1 \times 10^{-6}}{0.2}-\frac{2 \times 10^{-6}}{0.2}+\frac{3 \times 10^{-6}}{0.3}\right] \\
&=9 \times 10^{9}[5-10+10] \times 10^{-6} \\
&=9 \times 10^{3}[5]=45 \times 10^{3} \mathrm{~V}=45 \mathrm{kV}
\end{aligned}$
$\begin{aligned}
\mathrm{V} &=\frac{1}{4 \pi \in_{0}}\left[\frac{\mathrm{q}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{q}_{2}}{\mathrm{r}_{2}}+\frac{\mathrm{q}_{3}}{\mathrm{r}_{3}}\right] \\
\mathrm{V} &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{1 \times 10^{-6}}{0.2}-\frac{2 \times 10^{-6}}{0.2}+\frac{3 \times 10^{-6}}{0.3}\right] \\
&=9 \times 10^{9}[5-10+10] \times 10^{-6} \\
&=9 \times 10^{3}[5]=45 \times 10^{3} \mathrm{~V}=45 \mathrm{kV}
\end{aligned}$
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