Question

Three charges $-q$, $Q$ and $-q$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero then the ratio $\frac{Q}{q}$ is

• A. $1 : 2$
• B. $1 : 1$
• C. $1 : 4$
• D. $1 : 3$

Hint:

For symmetrical collinear charges, the total system energy can be calculated rapidly by focusing on a single side. The central charge has double the pulling interaction ($2 \times \frac{-qQ}{x}$), which must balance the long-distance repulsion of the outer two components ($\frac{q^2}{2x}$). Equating them instantly isolates $2Q = \frac{q}{2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Three point charges are aligned symmetrically along a straight line. The central charge is $Q$, flanked by two identical charges $-q$ on either side at an equal distance $x$. We need to find the value ratio $\frac{Q}{q}$ given that the net electrostatic potential energy of the entire system sums up to zero.

Step 2: Key Formula or Approach:
The electrostatic potential energy $U$ between two point charges $q_i$ and $q_j$ separated by a distance $r_{ij}$ is given by:
$$U = \frac{1}{4\pi\varepsilon_0} \frac{q_i q_j}{r_{ij}}$$ For a three-charge network, the total potential energy is the sum of the energies of all unique interacting pairs:
$$U_{\text{total}} = U_{12} + U_{23} + U_{13}$$

Step 3: Detailed Explanation:
Let's place the three charges along a coordinate axis:
Charge 1 (left): $-q$ at position $0$
Charge 2 (center): $Q$ at position $x$
Charge 3 (right): $-q$ at position $2x$
Let's calculate the individual pair-wise interactions:
1. Interaction between Charge 1 and Charge 2 (separated by distance $x$):
$$U_{12} = \frac{1}{4\pi\varepsilon_0} \frac{(-q)(Q)}{x}$$ 2. Interaction between Charge 2 and Charge 3 (separated by distance $x$):
$$U_{23} = \frac{1}{4\pi\varepsilon_0} \frac{(Q)(-q)}{x}$$ 3. Interaction between Charge 1 and Charge 3 (separated by distance $2x$):
$$U_{13} = \frac{1}{4\pi\varepsilon_0} \frac{(-q)(-q)}{2x} = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{2x}$$ Sum these values together and equate the total potential energy expression to zero:
$$\frac{1}{4\pi\varepsilon_0} \left[ \frac{-qQ}{x} + \frac{-qQ}{x} + \frac{q^2}{2x} \right] = 0$$ Since $\frac{1}{4\pi\varepsilon_0 x} \neq 0$, we can divide it out from the equation:
$$-2qQ + \frac{q^2}{2} = 0$$ Rearranging the terms:
$$2qQ = \frac{q^2}{2}$$ Divide both sides by $q$ (assuming $q \neq 0$):
$$2Q = \frac{q}{4} \implies 2Q = \frac{q}{2} \implies \frac{Q}{q} = \frac{1}{4}$$ Thus, the ratio $\frac{Q}{q}$ is $1:4$.

Step 4: Final Answer:
The ratio $\frac{Q}{q}$ is $1 : 4$, which corresponds to option (C).