Question

Four charges are placed very close to each other, as shown. The separation between the two charges on the y-axis is \( a \). The separation between the two charges on the x-axis is also \( a \). The leading order (non-vanishing) form of the electrostatic potential, at point \( P \), at a distance \( r \) from the origin (\( r \gg a \)), is: 

• A. \( \frac{1}{4\pi\varepsilon_0} \frac{qa}{2r^2}(\sqrt{3} - 1) \)
• B. \( \frac{1}{4\pi\varepsilon_0} \frac{2qa}{r^2} \)
• C. \( \frac{1}{4\pi\varepsilon_0} \frac{qa}{r^2}(\sqrt{5} - 1) \)
• D. \( \frac{1}{4\pi\varepsilon_0} \frac{qa}{r^2}(1 - \sqrt{3}) \)

Hint:

For charge systems with zero net charge, the dipole term gives the leading contribution to the potential at large distances.

Answer 1

The Correct Option is A

Step 1: Analyze the charge configuration.
The given system has four charges: \[ (+q \text{ at } +a \text{ on x-axis}), \quad (-q \text{ at } -a \text{ on x-axis}), \] \[ (+q \text{ at } +a \text{ on y-axis}), \quad (-q \text{ at } -a \text{ on y-axis}). \] The total charge is zero, so monopole contribution vanishes. The next term (dipole term) dominates at large distances.
Step 2: Calculate the net dipole moment.
Each pair contributes a dipole moment of magnitude \( 2qa \). The resultant dipole moment is along the bisector between x and y axes, at \( 45^\circ \), with magnitude \[ p = \sqrt{(2qa)^2 + (2qa)^2} = 2\sqrt{2}qa. \] Step 3: Potential due to dipole.
Potential at a point \( P \) far away (at an angle \( \theta \) from dipole axis): \[ V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2}. \] Given \( P \) makes a \( 60^\circ \) angle with the x-axis, the angle between \( P \) and dipole axis (which is \( 45^\circ \)) is \( \theta = 15^\circ \).
Step 4: Substitute values.
\[ V = \frac{1}{4\pi\varepsilon_0} \frac{2\sqrt{2}qa}{r^2} \cos(15^\circ). \] Using \( \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \), \[ V = \frac{1}{4\pi\varepsilon_0} \frac{qa}{2r^2}(\sqrt{3} - 1). \] Step 5: Final Answer.
Thus, the potential at \( P \) is \( \frac{1}{4\pi\varepsilon_0} \frac{qa}{2r^2}(\sqrt{3} - 1). \)