Question

Three charges +4q, Q and q are placed in a straight line of length $l$ at points at distance 0, $l/2$ and $l$ respectively from one end of line. What should be the value of Q in order to make the net force on q to be zero?

• A. -q
• B. -2q
• C. -q/2
• D. 4q

Answer 1

Answer: (A)

-q

Answer 2

Let the three charges +4q, Q, and q be placed on the x-axis at positions x=0, x=l/2, and x=l, respectively. We want to find the value of Q such that the net force on the charge q (at x=l) is zero.

The net force on q is the vector sum of the force on q due to +4q and the force on q due to Q.

1. Force on q due to +4q: The charge +4q is at x=0 and q is at x=l. The distance between them is $l$. Both charges are positive, so the force is repulsive and acts in the positive x direction (away from +4q). $F_{4q, q} = k \frac{(+4q)(q)}{l^2} = \frac{4kq^2}{l^2}$, where $k = \frac{1}{4\pi\epsilon_0}$. This force is in the positive x direction.

2. Force on q due to Q: The charge Q is at x=l/2 and q is at x=l. The distance between them is $l - l/2 = l/2$. $F_{Q, q} = k \frac{Q q}{(l/2)^2} = k \frac{Q q}{l^2/4} = \frac{4kQq}{l^2}$. The direction of this force depends on the sign of Q. If Q is positive, the force is repulsive (away from Q, in the positive x direction). If Q is negative, the force is attractive (towards Q, in the negative x direction).

For the net force on q to be zero, the forces $F_{4q, q}$ and $F_{Q, q}$ must be equal in magnitude and opposite in direction. Since $F_{4q, q}$ is in the positive x direction, $F_{Q, q}$ must be in the negative x direction. This requires Q to be a negative charge.

The net force on q is $F_{net, q} = F_{4q, q} + F_{Q, q}$ (treating forces as vectors along the x-axis). Let the positive x direction be the positive direction for forces. $F_{net, q} = +\frac{4kq^2}{l^2} + \frac{4kQq}{l^2}$. We set the net force to zero: $\frac{4kq^2}{l^2} + \frac{4kQq}{l^2} = 0$. Factor out the common term $\frac{4kq}{l^2}$: $\frac{4kq}{l^2}(q + Q) = 0$. Since $k \neq 0$, $q \neq 0$, and $l \neq 0$, we must have: $q + Q = 0$. $Q = -q$.

This value of Q is negative, which is consistent with our requirement that the force $F_{Q, q}$ must be attractive (in the negative x direction) to balance the repulsive force $F_{4q, q}$.