Question

Two charged particles of equal mass m and charges q, -2q are released at different location in uniform electric field $E_0$. If their relative acceleration is zero, then (Neglect the effect of gravity)

• A. Separation between particles is $\sqrt{\frac{q}{3\pi\epsilon_0E_0}}$
• B. Separation between particles is $\sqrt{\frac{2q}{3\pi\epsilon_0E_0}}$
• C. Acceleration of each particle is $\frac{qE_0}{2m}$
• D. Acceleration of each particle is $\frac{qE_0}{m}$

Answer 1

Answer: (A), (C)

Separation between particles is $\sqrt{\frac{q}{3\pi\epsilon_0E_0}}$ and Acceleration of each particle is $\frac{qE_0}{2m}

Answer 2

Let the particles have charges $q$ and $-2q$ (both of mass $m$). The net forces acting on the particles in the uniform electric field $E_0$ are:

For particle 1:

$$m\,a_1 = qE_0 + F_{12},\quad \text{where } F_{12} = \frac{k\,(q)(-2q)}{r^2} = -\frac{2kq^2}{r^2}.$$

So,

$$m\,a_1 = qE_0 - \frac{2kq^2}{r^2}.$$

For particle 2:

$$m\,a_2 = -2qE_0 + F_{21},\quad \text{with } F_{21} = \frac{k\,(-2q)(q)}{r^2} = -\frac{2kq^2}{r^2}.$$

But note that due to Newton's third law, the force on particle 2 due to particle 1 is actually in the opposite direction:

$$F_{21} = \frac{2kq^2}{r^2}.$$

Thus,

$$m\,a_2 = -2qE_0 + \frac{2kq^2}{r^2}.$$

The condition for zero relative acceleration is:

$$a_1 = a_2$$

Hence,

$$qE_0 - \frac{2kq^2}{r^2} = -2qE_0 + \frac{2kq^2}{r^2}.$$

Rearrange:

$$qE_0 + 2qE_0 = \frac{2kq^2}{r^2} + \frac{2kq^2}{r^2} \quad \Longrightarrow \quad 3qE_0 = \frac{4kq^2}{r^2}.$$

Cancel $q$ (assuming $q \neq 0$):

$$3E_0 = \frac{4kq}{r^2} \quad \Longrightarrow \quad r^2 = \frac{4kq}{3E_0}.$$

Since $k = \frac{1}{4\pi\epsilon_0}$,

$$r^2 = \frac{4\left(\frac{1}{4\pi\epsilon_0}\right)q}{3E_0} = \frac{q}{3\pi\epsilon_0E_0},$$ $$r = \sqrt{\frac{q}{3\pi\epsilon_0E_0}}.$$

Now, substitute $r^2$ back to find the acceleration. For particle 1:

$$m\,a_1 = qE_0 - \frac{2kq^2}{r^2}.$$

But,

$$\frac{2kq^2}{r^2} = \frac{2kq^2}{\frac{4kq}{3E_0}} = \frac{2kq^2 \cdot 3E_0}{4kq} = \frac{3qE_0}{2}.$$

Thus,

$$m\,a_1 = qE_0 - \frac{3qE_0}{2} = -\frac{qE_0}{2},$$ $$a_1 = -\frac{qE_0}{2m}.$$

The negative sign indicates the direction; the magnitude of the acceleration is $\frac{qE_0}{2m}$. Similarly, one finds the same magnitude for $a_2$.