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

Answer 1

Answer: (B)

Zero

Answer 2

Potential at A = Potential due to (+q) charge

+ Potential due to (– q) charge

$= \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{\sqrt{a^{2} + b^{2}}} + \frac{1}{4\pi\varepsilon_{0}}\frac{( - q)}{\sqrt{a^{2} + b^{2}}} = 0$