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)

$E\ne 0$ and $V=0$

Answer 2

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