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 $
BETA
