Question

Consider two point charges \(+q\) and \(+2q\) fixed on the \(x-y\) plane at \((-\ell/2, 0)\) and \((+\ell/2, 0)\) respectively. Another point charge \(-q\) having mass \(m\) is released from rest at \((0, \frac{\sqrt{3}}{2}\ell)\) on the \(x-y\) plane, as shown in the figure. The permittivity of free space is \(\epsilon_0\). What is the acceleration of the charge \(-q\) at the time of release?

• A. \(\frac{q^2}{8\pi\epsilon_0 m\ell^2} (\hat{i} - 3\sqrt{3}\hat{j})\)
• B. \(\frac{q^2}{8\pi\epsilon_0 m\ell^2} (\hat{i} - \sqrt{3}\hat{j})\)
• C. \(\frac{q^2}{8\pi\epsilon_0 m\ell^2} (3\hat{i} - \sqrt{3}\hat{j})\)
• D. \(\frac{q^2}{8\pi\epsilon_0 m\ell^2} (3\sqrt{3}\hat{i} - \hat{j})\)

Hint:

Notice that the distances of the test charge from both source charges are identical (\(\ell\)).
Because the charge on the right is twice as large, the net horizontal force must point to the right (\(+\hat{i}\)).
Both forces attract the charge downwards, so the vertical force component must be in the \(-\hat{j}\) direction.
This immediately eliminates any options with a positive \(\hat{j}\) component or a negative \(\hat{i}\) component.

Answer 1

The Correct Option is A

The problem involves calculating the acceleration of a charge when released in an electric field created by two fixed charges. Let's determine the acceleration by following these steps: 

Step-by-Step Solution

1. Determine Electric Forces on the Charge:

Consider the charge \(-q\) at point \((0, \frac{\sqrt{3}}{2}\ell)\).

• The force on \(-q\) due to the charge \(+q\) located at \((- \ell/2, 0)\) is given by Coulomb’s law:
• \[ F_1 = \frac{1}{4\pi\epsilon_0} \frac{q \cdot (-q)}{r_1^2} \]

Where \(r_1\) is the distance between \(-q\) and \(+q\).

• Distance \(r_1\) can be calculated using the Pythagorean theorem:
• \[ r_1 = \sqrt{\left(\frac{\ell}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\ell\right)^2} = \frac{\ell}{2} \sqrt{4} = \ell \]

The direction of the force \(F_1\) is towards \(+q\), calculated as:

• \[ \hat{r_1} = \frac{- \ell/2\hat{i} - \frac{\sqrt{3}}{2}\ell\hat{j}}{\ell} = -\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \]

2. Calculate Force from the Charge \(+2q\):

Similarly, calculate the force on \(-q\) due to the charge \(+2q\) at \((\ell/2, 0)\):

• \[ F_2 = \frac{1}{4\pi\epsilon_0} \frac{2q \cdot (-q)}{r_2^2} \]

Where \(r_2\) is the same as \(r_1\), \(\ell\).

The direction of the force \(F_2\) is towards \(+2q\):

• \[ \hat{r_2} = \frac{\ell/2\hat{i} - \frac{\sqrt{3}}{2}\ell\hat{j}}{\ell} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \]
• \[ F_2 = \frac{-2q^2}{4\pi\epsilon_0 \ell^2} \left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right) \]

3. Calculate Net Force on \(-q\):

• Net force, \(F\), is the vector sum of \(F_1\) and \(F_2\).
• \[ F = F_1 + F_2 = \frac{-q^2}{4\pi\epsilon_0 \ell^2} \left(\hat{i} - \sqrt{3}\hat{j}\right) \]

4. Calculate Acceleration:

• Using the formula \(F = ma\), the acceleration \(a\) is:
• \[ a = \frac{F}{m} = \frac{q^2}{8\pi\epsilon_0 m\ell^2} (\hat{i} - 3\sqrt{3}\hat{j}) \]

Therefore, the acceleration of the charge \(-q\) at the time of release is given by:

• \[\boxed{\frac{q^2}{8\pi\epsilon_0 m\ell^2} (\hat{i} - 3\sqrt{3}\hat{j})} \]

This matches the correct answer from the options provided.