Question

Two positive charges of +Q each are placed at the corners A and B of an equilateral triangle ABC of side R. The net electric potential at C, in terms of K is (\(K = \frac{Q}{4\pi\epsilon_0 R}\))

• A. \(\sqrt{2}K\)
• B. \(\sqrt{3}K\)
• C. 2K
• D. K
• E. \(\sqrt{5}K\)

Hint:

Potentials add up directly as numbers (scalars), whereas electric fields add up as vectors. Never use vector addition (like \(\sqrt{E_A^2+E_B^2...}\)) for potential problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Electric potential is a scalar quantity. The total potential at a point due to multiple charges is the algebraic sum of the potentials due to each individual charge.

Step 2: Key Formula or Approach:
Potential \(V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r}\).

Step 3: Detailed Explanation:
1. Both charges are placed at distance \(R\) from point C.
2. Potential at C due to charge at A:
\[ V_A = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} \]
3. Potential at C due to charge at B:
\[ V_B = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{R} \]
4. Total potential at C:
\[ V_{net} = V_A + V_B = \frac{Q}{4\pi\epsilon_0 R} + \frac{Q}{4\pi\epsilon_0 R} = 2 \cdot \left( \frac{Q}{4\pi\epsilon_0 R} \right) \]
5. Using the given definition \(K = \frac{Q}{4\pi\epsilon_0 R}\):
\[ V_{net} = 2K \]

Step 4: Final Answer:
The net electric potential at C is 2K.