Question

A point charge of +2 nC is kept at the origin of a three-dimensional coordinate system. Find the type and magnitude of the charge which should be kept at (0, 0, −6 m) so that the potential due to the system becomes zero at (0, 0, 2 m).

Answer 1

Objective: Determine Unknown Charge from Electric Potential

Given Data:

• Charge \( q_1 = +2 \, \text{nC} = 2 \times 10^{-9} \, \text{C} \) located at \( (0, 0, 0) \)
• Unknown charge \( q_2 \) is at \( (0, 0, -6) \, \text{m} \)
• Observation point: \( (0, 0, 2) \, \text{m} \)
• The net electric potential at \( (0, 0, 2) \, \text{m} \) is 0.

Methodology:

Step 1: Electric Potential Formula

The electric potential \( V \) generated by a point charge \( q \) at a distance \( r \) is given by:

\[ V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r} \]

Step 2: Calculate Distances to Observation Point

• Distance from \( q_1 \) to the observation point: \( r_1 = 2 \, \text{m} \)
• Distance from \( q_2 \) to the observation point: \( r_2 = |2 - (-6)| \, \text{m} = 8 \, \text{m} \)

Step 3: Sum of Potentials Calculation

The total electric potential at the point is the superposition of the potentials from individual charges:

\[ V_{\text{total}} = V_1 + V_2 = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right) = 0 \]

Substituting known values:

\[ \frac{1}{4\pi\varepsilon_0} \left( \frac{2 \times 10^{-9} \, \text{C}}{2 \, \text{m}} + \frac{q_2}{8 \, \text{m}} \right) = 0 \]

Simplifying the equation:

\[ 1 \times 10^{-9} + \frac{q_2}{8} = 0 \]

Solving for \( q_2 \):

\[ \frac{q_2}{8} = -1 \times 10^{-9} \, \text{C} \]

\[ q_2 = -8 \times 10^{-9} \, \text{C} \]

Conclusion:

The unknown charge is: \( q_2 = -8 \, \text{nC} \), situated at \( (0, 0, -6) \, \text{m} \)