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)
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)
(Potential at infinity is assumed to be zero)
Show 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 \(x\).
Updated On: Jun 19, 2026
- \(\frac{8}{3}~\text{cm}\)
- \(\frac{4}{3}~\text{cm}\)
- \(\frac{2}{3}~\text{cm}\)
- \(\frac{16}{3}~\text{cm}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Condition for zero potential.
At a point \(x\) from \(q_1\) along the line joining the charges: \[ V = k \frac{q_1}{x} + k \frac{q_2}{d - x} = 0 \] where \(d = 8~\text{cm}\).
Step 2: Substitute values.
\[ \frac{4 \times 10^{-6}}{x} + \frac{-2 \times 10^{-6}}{8 - x} = 0 \]
Step 3: Solve for \(x\).
\[ \frac{4}{x} - \frac{2}{8 - x} = 0 \Rightarrow \frac{4}{x} = \frac{2}{8 - x} \Rightarrow 4 (8 - x) = 2 x \] \[ 32 - 4x = 2x \Rightarrow 32 = 6x \Rightarrow x = \frac{32}{6} = \frac{16}{3}~\text{cm} \]
Step 4: Conclusion.
The point where the potential vanishes is \(\frac{16}{3}~\text{cm}\) from \(q_1\).
At a point \(x\) from \(q_1\) along the line joining the charges: \[ V = k \frac{q_1}{x} + k \frac{q_2}{d - x} = 0 \] where \(d = 8~\text{cm}\).
Step 2: Substitute values.
\[ \frac{4 \times 10^{-6}}{x} + \frac{-2 \times 10^{-6}}{8 - x} = 0 \]
Step 3: Solve for \(x\).
\[ \frac{4}{x} - \frac{2}{8 - x} = 0 \Rightarrow \frac{4}{x} = \frac{2}{8 - x} \Rightarrow 4 (8 - x) = 2 x \] \[ 32 - 4x = 2x \Rightarrow 32 = 6x \Rightarrow x = \frac{32}{6} = \frac{16}{3}~\text{cm} \]
Step 4: Conclusion.
The point where the potential vanishes is \(\frac{16}{3}~\text{cm}\) from \(q_1\).
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