Question

As shown in the figure, charges $+ q$ and $- q$ are placed at the vertices $B$ and $C$ of an isosceles triangle. The potential at the vertex $A$ is

• A. $\frac{1}{4\pi\varepsilon_{0}}. \frac{2a}{\sqrt{a^{2}+b^{2}}}$
• B. zero
• C. $\frac{1}{4\pi\varepsilon_{0}}. \frac{q}{\sqrt{a^{2}+b^{2}}}$
• D. $\frac{1}{4\pi\varepsilon_{0}}. \frac{\left(-q\right)}{\sqrt{a^{2}+b^{2}}}$

Answer 1

Answer: (B)

zero

Answer 2

Potential at $A$
$=+\frac{q}{\sqrt{a^{2}+b^{2}}}-\frac{q}{\sqrt{a^{2}+b^{2}}}$
= zero