Electrostatics Question 416

Question:charges $ +q $ and $ -q $ are placed at the vertices $ B $ and $ C $ of an isosceles triangle. The potential at the vertex A is [MP PET 2000]

Options:

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}}} $

Show Answer

Answer:

Correct Answer: B

Solution:

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 $



NCERT Chapter Video Solution

Dual Pane