![Q.21 Two charges 10 × 10-8 C and - 6 ×10-8 C are located 16 cm apart. The point(s) on the line joining the two charges, where the net electric potential is zero, will be [Take the potential at infinity to be zero] (A) 10 cm and 40 cm away from the positive charge on the side of the negative charge (B) 10 cm and 40 cm away from the negative charge on the side of the positive charge (C) 10 cm away from the negative charge on the side of the positive charge (D) 40 cm away from the negative charge on the side of the positive charge](data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
Q.21
Two charges 10 × 10-8 C and – 6 ×10-8 C are located 16 cm apart. The point(s) on the line
joining the two charges, where the net electric potential is zero, will be
[Take the potential at infinity to be zero]
(A) 10 cm and 40 cm away from the positive charge on the side of the negative charge
(B) 10 cm and 40 cm away from the negative charge on the side of the positive charge
(C) 10 cm away from the negative charge on the side of the positive charge
(D) 40 cm away from the negative charge on the side of the positive charge
Electric Potential Zero Between Two Charges: Solved with Explanation
Two charges of \(10 \times 10^{-8}\) C and \(-6 \times 10^{-8}\) C separated by 16 cm have specific points where net electric potential is zero, crucial for JEE and physics exam prep.
Problem Setup
The positive charge \(q_1 = 10 \times 10^{-8}\) C (or \(1 \times 10^{-7}\) C) and negative charge \(q_2 = -6 \times 10^{-8}\) C lie 16 cm apart on a line. Electric potential \(V\) at a point is \(V = k \frac{q_1}{r_1} + k \frac{q_2}{r_2}\), where \(k = \frac{1}{4\pi\epsilon_0}\), and zero at infinity. Set \(V = 0\) to find points: \(\frac{q_1}{r_1} = -\frac{q_2}{r_2}\) or \(\frac{10 \times 10^{-8}}{r_1} = \frac{6 \times 10^{-8}}{r_2}\).
Since \(|q_1| > |q_2|\), the null point closer to \(q_2\) (negative) balances potentials; farther point exists beyond \(q_2\).
Correct Answer: Option (A)
Points are 10 cm and 40 cm from positive charge, on negative charge side (between and beyond).
Let positive charge at x=0, negative at x=16 cm.
\(\frac{10^{-7}}{x} = \frac{6 \times 10^{-8}}{16 – x}\) → \(10(16 – x) = 6x\) → \(160 = 16x\) → \(x = 10\) cm.
\(\frac{10^{-7}}{16 + s} = \frac{6 \times 10^{-8}}{s}\) → \(10s = 6(16 + s)\) → \(10s = 96 + 6s\) → \(4s = 96\) → \(s = 24\) cm, so from positive: 40 cm.
Option Analysis
| Option | Description | Why Incorrect/Correct |
|---|---|---|
| (A) | 10 cm & 40 cm from +ve on -ve side | Correct. Matches calculations: 10 cm between, 40 cm beyond -ve charge. |
| (B) | 10 cm & 40 cm from -ve on +ve side | Incorrect. On +ve side (left of +ve), distances don’t satisfy ratio 10:6; potential dominated by +ve. |
| (C) | Only 10 cm from -ve on +ve side | Incorrect. Ignores outer point; 10 cm from -ve toward +ve is x=6 cm from +ve, but equation gives x=10 cm. |
| (D) | Only 40 cm from -ve on +ve side | Incorrect. Single point wrong; 40 cm from -ve on +ve side exceeds separation without zero. |
No points on positive side due to charge magnitudes.
