Question

Two-point charge -q and +q/2 are situated at the origin and at the point (a, 0, 0) respectively. The point along the x-axis where the electric field vanishes is:

• A. x = a/√2
• B. x = $\sqrt{2}a$
• C. x = $\frac{\sqrt{2}a}{\sqrt{2}-1}$
• D. x = $\frac{\sqrt{2}a}{\sqrt{2}+1}$

Answer 1

Answer: (C)

x = $\frac{\sqrt{2}a}{\sqrt{2}-1}$

Answer 2

Let the point along the x-axis be at $x$ (with $x > a$) where the net electric field is zero.

1. Field due to $-q$ at the origin:

   $$E_{-q} = \frac{kq}{x^2}$$

   (Direction: towards the negative charge, i.e. leftwards for $x>a$).

2. Field due to $+\frac{q}{2}$ at $x=a$:

   $$E_{+q/2} = \frac{k\frac{q}{2}}{(x-a)^2}$$

   (Direction: away from the positive charge, i.e. rightwards for $x>a$).

In region $x>a$, the fields are oppositely directed, hence for equilibrium:

$$\frac{k\frac{q}{2}}{(x-a)^2} = \frac{kq}{x^2}$$

Cancel $k$ and $q$:

$$\frac{1}{2(x-a)^2} = \frac{1}{x^2} \quad \Longrightarrow \quad x^2 = 2(x-a)^2$$

Taking positive square roots (since $x > a$):

$$\frac{x}{x-a} = \sqrt{2} \quad \Longrightarrow \quad x = \sqrt{2}(x-a)$$

Solve for $x$:

$$x = \sqrt{2}x - \sqrt{2}a \quad \Longrightarrow \quad x( \sqrt{2}-1)= \sqrt{2}a \quad \Longrightarrow \quad x=\frac{\sqrt{2}a}{\sqrt{2}-1}$$