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

Answer 1

Answer: (D)

Zero

Answer 2

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

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