Question

Two charges \(10 \times 10^{-8}\,\text{C}\) and \(-6 \times 10^{-8}\,\text{C}\) 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

Hint:

For zero potential points, solve \(\frac{q_1}{r_1} + \frac{q_2}{r_2} = 0\). Both “between” and “outside” regions should be checked.

Answer 1

The Correct Option is A

Step 1: Recall formula.
Potential at a point due to a charge \(q\): \[ V = \frac{kq}{r} \] Step 2: Define the setup.
- \(q_1 = +10 \times 10^{-8}\,\text{C}\), placed at \(x=0\).
- \(q_2 = -6 \times 10^{-8}\,\text{C}\), placed at \(x=16\) cm.
We need points where \(V_1 + V_2 = 0\).
Step 3: Between the charges.
At distance \(x\) from \(+q_1\): \[ \frac{10}{x} = \frac{6}{16-x} \] \[ 10(16-x) = 6x \quad \Rightarrow \quad 160 - 10x = 6x \quad \Rightarrow \quad 16x = 160 \quad \Rightarrow \quad x=10\ \text{cm} \] Step 4: Outside the charges (on right side of -q).
At distance \(d\) from \(+q_1\) beyond the negative charge (\(x>16\)): \[ \frac{10}{d} = \frac{6}{d-16} \] \[ 10(d-16) = 6d \quad \Rightarrow \quad 10d - 160 = 6d \quad \Rightarrow \quad 4d = 160 \quad \Rightarrow \quad d = 40\ \text{cm} \] Step 5: Conclusion.
Zero potential points are at 10 cm and 40 cm from the positive charge on the side of the negative charge.
Final Answer: \[ \boxed{\text{10 cm and 40 cm away from the positive charge on the side of the negative charge.}} \]