Question

Charges $2q,-q$ and $-q$ lie at the vertices of a triangle. The value of $E$ and Vat the centroid of equilateral triangle will be

• A. $E\ne 0$ and $V\ne 0$
• B. $E=0$ and $V=0$
• C. $E\ne 0$ and $V=0$
• D. $E=0$ and $V\ne 0$

Answer 1

Answer: (C)

The potential due to charge $q$ at a distance $r$ is given by
$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
Since, potential is a scalar quantity, it can be added to find the sum due to individual charges.
$\Sigma V=V_A+V_B+V_C{V}_A$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2q}{x}$
$V_B=-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{x}$
$V_C=-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{x}$
$\therefore V=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{x}-\frac{q}{x}-\frac{q}{x}\right)=0$
Electric field is a vector quantity, hence component along $OD$ is taken
$E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{x^2}+\frac{2q}{x^2}\cos \theta \right)\ne 0$