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

Step 1: Understanding the Question:
We need to find the acceleration vector of a charge \(-q\) placed at point \(P(0, \frac{\sqrt{3}}{2}\ell)\) due to two fixed positive charges: \(+q\) at \(A(-\ell/2, 0)\) and \(+2q\) at \(B(\ell/2, 0)\).

Step 2: Key Formula or Approach:

• Coulomb's Law:
\[ \vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r} \]
• Superposition Principle:
\[ \vec{F}_{\text{total}} = \vec{F}_A + \vec{F}_B \]
• Acceleration:
\[ \vec{a} = \frac{\vec{F}_{\text{total}}}{m} \]

Step 3: Detailed Explanation:

• Distance from \(A(-\ell/2, 0)\) to \(P(0, \frac{\sqrt{3}}{2}\ell)\).:
\[ r_{AP} = \sqrt{\left(0 - \left(-\frac{\ell}{2}\right)\right)^2 + \left(\frac{\sqrt{3}}{2}\ell - 0\right)^2} = \sqrt{\frac{\ell^2}{4} + \frac{3\ell^2}{4}} = \ell \]
• Similarly, distance from \(B(\ell/2, 0)\) to \(P(0, \frac{\sqrt{3}}{2}\ell)\).:
\[ r_{BP} = \sqrt{\left(0 - \frac{\ell}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\ell - 0\right)^2} = \ell \]
• The unit vector from \(A\) to \(P\) is:
\[ \hat{r}_{AP} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \]
• The unit vector from \(B\) to \(P\) is:
\[ \hat{r}_{BP} = -\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \]
• Force on \(-q\) due to \(+q\) at \(A\).:
\[ \vec{F}_A = \frac{1}{4\pi\epsilon_0} \frac{q(-q)}{\ell^2} \hat{r}_{AP} = -\frac{q^2}{4\pi\epsilon_0\ell^2} \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) \]
• Force on \(-q\) due to \(+2q\) at \(B\).:
\[ \vec{F}_B = \frac{1}{4\pi\epsilon_0} \frac{2q(-q)}{\ell^2} \hat{r}_{BP} = -\frac{2q^2}{4\pi\epsilon_0\ell^2} \left(-\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) \]
• Adding these forces:
\[ \vec{F}_{\text{total}} = -\frac{q^2}{4\pi\epsilon_0\ell^2} \left[ \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) + 2\left(-\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) \right] \] \[ \vec{F}_{\text{total}} = -\frac{q^2}{4\pi\epsilon_0\ell^2} \left[ -\frac{1}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j} \right] = \frac{q^2}{8\pi\epsilon_0\ell^2} \left[ \hat{i} - 3\sqrt{3}\hat{j} \right] \]
• The acceleration is:
\[ \vec{a} = \frac{\vec{F}_{\text{total}}}{m} = \frac{q^2}{8\pi\epsilon_0 m \ell^2} (\hat{i} - 3\sqrt{3}\hat{j}) \]

Step 4: Final Answer:
The acceleration of the charge is \(\frac{q^2}{8\pi\epsilon_0 m \ell^2} (\hat{i} - 3\sqrt{3}\hat{j})\).