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: Arrange the charges.
Place $-q$, $Q$, $-q$ in a line with equal spacing $a$: $-q$ at one end, $Q$ in the middle, $-q$ at the other end. The two end charges are then $2a$ apart. We need $\dfrac{Q}{q}$ so the total potential energy is zero.
Step 2: Recall the pair energy.
Each pair contributes $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i q_j}{r_{ij}}$, and the system energy is the sum over all three pairs.
Step 3: Middle-to-end pairs.
The two $Q$-with-$(-q)$ pairs are each at distance $a$: each gives $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-qQ}{a}$. Together they give $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-2qQ}{a}$.
Step 4: End-to-end pair.
The two $-q$ charges are $2a$ apart, giving $\dfrac{1}{4\pi\varepsilon_0}\dfrac{(-q)(-q)}{2a} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{2a}$.
Step 5: Set the total to zero.
$\dfrac{1}{4\pi\varepsilon_0 a}\left(-2qQ + \dfrac{q^2}{2}\right) = 0$. The prefactor is non-zero, so $-2qQ + \dfrac{q^2}{2} = 0$.
Step 6: Solve the ratio.
Then $2qQ = \dfrac{q^2}{2}$, giving $Q = \dfrac{q}{4}$, so $\dfrac{Q}{q} = \dfrac{1}{4}$, i.e. $1 : 4$, option (3).
\[ \boxed{\dfrac{Q}{q} = \dfrac{1}{4}\ (1:4)} \]