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

Problem: Finding Unknown Charge Using Electric Potential

Given:

• \( q_1 = +2 \, \text{nC} = 2 \times 10^{-9} \, \text{C} \) at position \( (0, 0, 0) \)
• \( q_2 \) is an unknown charge at position \( (0, 0, -6) \, \text{m} \)
• Point of interest: \( (0, 0, 2) \, \text{m} \)
• Total potential at point \( (0, 0, 2) \) is zero.

Step 1: Formula for Electric Potential

Electric potential due to a point charge at a distance \( r \) is:

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

Step 2: Distances from Point of Interest

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

Step 3: Total Electric Potential

The total potential at the point is the sum of potentials from both charges:

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

Substitute values:

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

Final Answer:

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