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. \( \frac{1}{4\pi\epsilon_0} \frac{2a}{\sqrt{a^2 + b^2}} \)
• B. \( \frac{1}{4\pi\epsilon_0} \frac{2a}{\sqrt{a^2 + b^2}} \)
• C. \( \frac{1}{4\pi\epsilon_0} \frac{-2a}{\sqrt{a^2 + b^2}} \)
• D. \( \frac{1}{4\pi\epsilon_0} \frac{2a}{\sqrt{a^2 + b^2}} \)

Hint:

The electric potential at a point due to multiple charges is the algebraic sum of the potentials due to individual charges.

Answer 1

The Correct Option is A

Step 1: Use the formula for potential.
The potential at a point due to a charge \( q \) is given by: \[ V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r} \] where \( r \) is the distance from the charge.
Step 2: Calculation.
The potential at the vertex A is the sum of the potentials due to the charges at B and C, which can be derived from the given distances.
Final Answer: \[ \boxed{\frac{1}{4\pi\epsilon_0} \frac{2a}{\sqrt{a^2 + b^2}}} \]