Question

Two charges \( +5 \, \mu\text{C} \) and \( +5 \, \mu\text{C} \) are placed 1 m apart. What is the electric potential at the midpoint between them? (Take \(k = 9 \times 10^9 \, \text{N.m}^2/\text{C}^2\)).

• A. \( 9 \times 10^4 \, \text{V} \)
• B. \( 1.8 \times 10^5 \, \text{V} \)
• C. \( 2.7 \times 10^5 \, \text{V} \)
• D. \( 3.6 \times 10^5 \, \text{V} \)

Hint:

For electric potential at a point due to multiple charges, sum the potentials \( V = k \frac{q}{r} \) for each charge, using consistent units (charge in C, distance in m).

Answer 1

The Correct Option is B

The electric potential \( V \) due to a point charge \( q \) at distance \( r \) is: \[ V = k \frac{q}{r} \] The charges are \( q_1 = q_2 = 5 \times 10^{-6} \, \text{C} \), and they are 1 m apart, so the midpoint is at \( r = 0.5 \, \text{m} \) from each charge. The total potential at the midpoint is the sum of potentials due to both charges: \[ V_{\text{total}} = V_1 + V_2 = k \frac{q_1}{r} + k \frac{q_2}{r} = \frac{k}{r} (q_1 + q_2) \] \[ = \frac{9 \times 10^9}{0.5} (5 \times 10^{-6} + 5 \times 10^{-6}) = 18 \times 10^9 \cdot 10 \times 10^{-6} = 18 \times 10^4 = 1.8 \times 10^5 \, \text{V} \] The electric potential is: \[ \boxed{1.8 \times 10^5} \]