Question

Three charges -q, Q and -q are placed in a straight line maintaining equal distance from each other. What should be the ratio \(\frac{q}{Q}\) so that net electric potential of the system is zero?

Answer 1

Correct Answer: 4

Understanding the Problem:

• Three point charges are arranged linearly: \(-q\), \(Q\), and \(-q\)
• Equal spacing \(d\) between consecutive charges
• Find the ratio \(\frac{q}{Q}\) for zero net electric potential energy

Key Formulae:

• Coulomb's law: \(V = k\frac{q}{r}\) where \(k = \frac{1}{4\pi\epsilon_0}\)
• Potential energy between two charges: \(U = k\frac{q_1q_2}{r}\)

System Configuration:

• Charge arrangement: \(-q\) (left), \(Q\) (center), \(-q\) (right)
• Distances: \(-q\) to \(Q = d\), \(Q\) to \(-q = d\), \(-q\) to \(-q = 2d\)

Potential Energy Calculation:

Total potential energy \(U_{\text{total}}\) has three components:

1. Between \(-q\) and \(Q\): \[ U_1 = k\frac{(-q)(Q)}{d} = -k\frac{qQ}{d} \]

2. Between \(Q\) and \(-q\): \[ U_2 = k\frac{(Q)(-q)}{d} = -k\frac{qQ}{d} \]

3. Between \(-q\) and \(-q\): \[ U_3 = k\frac{(-q)(-q)}{2d} = k\frac{q^2}{2d} \]

Total Potential Energy:

\[ U_{\text{total}} = U_1 + U_2 + U_3 = -2k\frac{qQ}{d} + k\frac{q^2}{2d} \]

Condition for Zero Energy:

Set \(U_{\text{total}} = 0\): \[ -2k\frac{qQ}{d} + k\frac{q^2}{2d} = 0 \]

Simplify: \[ -4qQ + q^2 = 0 \] \[ q(q - 4Q) = 0 \]

Non-Trivial Solution:

\[ q - 4Q = 0 \] \[ \frac{q}{Q} = 4 \]

Final Answer: The required ratio is \(\boxed{\dfrac{q}{Q} = 4}\).