Question

Two point charges \( 20 \, \mu C \) and \( -10 \, \mu C \) are separated by a distance of 1 m in air. At what point on the line joining the two charges, the electric potential is zero.

• A. 0.5 m from the charge \( 10 \, \mu C \)
• B. 0.76 m from the charge \( 20 \, \mu C \)
• C. 0.67 m from the charge \( 20 \, \mu C \)
• D. 0.25 m from the charge \( 10 \, \mu C \)

Hint:

When the electric potential is zero due to two point charges, the magnitudes of the potentials due to both charges must be equal but opposite in sign. Set the sum of the potentials equal to zero and solve for the distance.

Answer 1

The Correct Option is C

The electric potential at a point due to a point charge is given by the formula: \[ V = \frac{kQ}{r} \] Where: - \( V \) is the electric potential, - \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), - \( Q \) is the charge, - \( r \) is the distance from the charge. At the point where the potential is zero, the potentials due to both charges must cancel each other out. Therefore, the total electric potential at that point is zero: \[ V_1 + V_2 = 0 \] For the two charges: - The charge \( Q_1 = 20 \, \mu C \) is at a distance \( r_1 \) from the point where the potential is zero. - The charge \( Q_2 = -10 \, \mu C \) is at a distance \( r_2 = 1 - r_1 \) from the same point. The electric potentials at the point due to the charges are: \[ \frac{k \times 20 \times 10^{-6}}{r_1} + \frac{k \times (-10 \times 10^{-6})}{(1 - r_1)} = 0 \] Simplifying this equation: \[ \frac{20 \times 10^{-6}}{r_1} = \frac{10 \times 10^{-6}}{(1 - r_1)} \] \[ \frac{20}{r_1} = \frac{10}{1 - r_1} \] Cross-multiply to solve for \( r_1 \): \[ 20(1 - r_1) = 10r_1 \] \[ 20 - 20r_1 = 10r_1 \] \[ 20 = 30r_1 \] \[ r_1 = \frac{20}{30} = 0.67 \, \text{m} \] Thus, the electric potential is zero at a point 0.67 m from the \( 20 \, \mu C \) charge.