Question

Three point charges $4q$, $Q$ and $q$ are placed in a straight line of length $L$ at points 0, $L/2$ and $L$ respectively. The net force on charge $q$ is zero. The value of $Q$ is

• A. $4q$
• B. $-q$
• C. $-0.5q$
• D. $-2q$
• E. $3q$

Hint:

If a charge is in equilibrium under the influence of two other charges of the same sign, the middle charge must have an opposite sign to provide an attractive force that cancels the repulsive one.

Answer 1

The Correct Option is B

Concept:
According to Coulomb's Law, the force between two charges $q_1$ and $q_2$ separated by distance $r$ is $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}$. For the net force on a charge to be zero, the vector sum of forces from all other charges must be zero.

Step 1: Identify the forces acting on charge $q$ (at $x=L$).
[label=\alph*), itemsep=8pt]
• Force from $4q$ (at $x=0$): $F_1 = \frac{1}{4\pi\varepsilon_0} \frac{(4q)(q)}{L^2}$ (Repulsive, points to the right).
• Force from $Q$ (at $x=L/2$): $F_2 = \frac{1}{4\pi\varepsilon_0} \frac{(Q)(q)}{(L/2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{4Qq}{L^2}$.

Step 2: Equate net force to zero.
\[ F_1 + F_2 = 0 \implies \frac{1}{4\pi\varepsilon_0} \frac{4q^2}{L^2} + \frac{1}{4\pi\varepsilon_0} \frac{4Qq}{L^2} = 0 \] \[ 4q^2 + 4Qq = 0 \implies 4q(q + Q) = 0 \] Since $q \neq 0$, $q + Q = 0 \implies Q = -q$.