Question

Two charges \(q_1 = 4 \times 10^{-6}~\text{C}\) and \(q_2 = -2 \times 10^{-6}~\text{C}\) are separated by 8 cm. The distance of the point from \(q_1\) on the line joining the charges where the potential vanishes is:
(Potential at infinity is assumed to be zero)

• A. \(\frac{8}{3}~\text{cm}\)
• B. \(\frac{4}{3}~\text{cm}\)
• C. \(\frac{2}{3}~\text{cm}\)
• D. \(\frac{16}{3}~\text{cm}\)

Hint:

For zero potential along the line joining two point charges, set algebraic sum of potentials to zero: \(\frac{q_1}{x} + \frac{q_2}{d - x} = 0\) and solve for \(x\).

Answer 1

The Correct Option is D

Step 1: Potential superposition condition.
At a distance x from q₁ along the line, V = k q₁/x + k q₂/(d-x) = 0, with d = 8 cm.

Step 2: Inserting charge values.
(4×10⁻⁶)/x + (-2×10⁻⁶)/(8-x) = 0.

Step 3: Solving the equation.
4/x = 2/(8-x) → 4(8-x) = 2x → 32 - 4x = 2x → 32 = 6x → x = 32/6 = 16/3 cm.

Step 4: Conclusion.
The potential becomes zero at 16/3 cm from q₁.