Question

Three identical charges, each with a value of 1.0 × 10⁻⁸ C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the centre of the triangle.

Answer 1

Given:
Magnitude of charges, q = 1.0 × 10⁻⁸ C
Side of the triangle, $l=20\mathrm{cm}=0.2\mathrm{m}$

Let $E_A,E_B\mathrm{and}{E}_C$be the electric fields at the centre due to the charges at A, B and C, respectively.
The distance of the centre is the same from all the charges. So,
$E_A=E_B=E_C=E\left(\mathrm{say}\right)$
Resolving $E_B\mathrm{and}{E}_C$ into vertical and horizontal components (with $\theta =30°$)
The horizontal components cancel each other, as shown in the figure.
So, the net electric field at the centre,
$$\begin{gathered} E_{net}=E_A-\left(E_B\sin \theta +E_A\sin \theta \right) \\ =E\left(1-\sin 30°-\sin 30°\right) \\ \Rightarrow E_{net}=0 \end{gathered}$$
Now,
$$\begin{gathered} h^2=l^2-{\left(\frac{l}{2}\right)}^2 \\ {h}^2={\left(0.2\right)}^2-{\left(0.1\right)}^2 \\ \Rightarrow h=\frac{\sqrt{3}}{10} \end{gathered}$$
Let the distance of the centre from each charge be r.
For an equilateral triangle,
$$\begin{gathered} r=\frac{2}{3}h \\ \Rightarrow r=\frac{2}{3}\times \frac{1.732}{10}=1.15\times {10}^{-1}\mathrm{m} \end{gathered}$$
Potential at the centre,
$$\begin{gathered} V=V_A+V_B+V_C \\ \because V_A=V_B=V_{C,} \\ V=3V_A \end{gathered}$$
$$\begin{gathered} V=3\times \frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q}{r} \\ V=\frac{3\times 9\times {10}^9\times {10}^{-8}}{0.115} \end{gathered}$$
$V=23\times {10}^2=2.3\times {10}^3\mathrm{V}$