Question

The side of an equilateral triangle is $100cm$. The charges at its three corners are $1\mu C,2\mu C$ and $3\mu C$ respectively. Calculate the electric potential at a point (in the centre) equidistant from the three corners ofthe triangle.

Answer 1

Given $a=100cm$
$=100\times {10}^{-2}m$
Draw a $\perp OD$ and $AB$ in $\Delta ODB$
$\begin{array}{c}\cos \left({30}^{\circ }\right)=\frac{DB}{OB}=\frac{a/2}{r}\\ r=\frac{a}{2\cos {30}^{\circ }}\end{array}$
$r=\frac{a}{2\times \frac{\sqrt{3}}{2}}\Rightarrow r=\frac{a}{\sqrt{3}}$
$\begin{array}{cc}{q}_1& =1\mu C\\ q_1& =1\times {10}^{-6}C\\ q_2& =2\mu C\\ & =2\times {10}^{-6}C\\ q_3& =3\mu C\\ & =3\times {10}^{-6}C\end{array}$
$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\frac{N-m^2}{C^2}$
Electric potential at centre of square
$\begin{array}{c}{V}_0=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{r}+\frac{q_2}{r}+\frac{q_3}{r}]\\ V_0=\frac{1}{4\pi {\epsilon }_0}\frac{1}{r}[1\times {10}^{-6}+2\times {10}^{-6}+3\times {10}^{-6}]\\ V_0=\frac{1\times {10}^{-6}}{4\pi {\epsilon }_0\times \frac{a}{\sqrt{3}}}\left[6\right]\\ V_0=\frac{9\times {10}^9\times {10}^{-6}\times 6\times \sqrt{3}}{100\times {10}^{-2}}\\ V_0=54\sqrt{3}\times {10}^3\\ \Rightarrow V_0=54\times 1.732\times {10}^3\Rightarrow V_0=93.6\times {10}^{3}V\end{array}$