Question

Four electric charges \(+q\) , \(+q\) , \(-q\) , and \(-q\) are placed in order at the corners of a square of side '\(2r\)'. The electric potential at a point midway between the two negative charges is

• A. \(\frac{1}{4πε_0}\,\frac{2q}{r}[\frac{1}{\sqrt{5}}-1]\)
• B. \(\frac{1}{4πε_0}\,\frac{q}{r}[\frac{1}{\sqrt{5}}+1]\)
• C. \(\frac{1}{4πε_0}\,\frac{2q}{r}[1-\sqrt{5}]\)
• D. \(\frac{1}{4πε_0}\,\frac{q}{r}[1+\sqrt{5}]\)

Hint:

Add the potentials of the four charges at the midpoint of one side; distances are r and r root 5.

Answer 1

The Correct Option is A

Step 1: Understanding the Figure:
The charges \(+q, +q, -q, -q\) are placed in order at the corners of a square of side \(2r\). The two negative charges are at adjacent corners, so the midpoint \(P\) of the side joining them is at distance \(r\) from each.

Step 2: Distances to the positive charges:
Each positive charge sits at the corner adjacent to one negative charge, across the side. From \(P\), the distance to such a corner is \(\sqrt{(2r)^2 + r^2} = r\sqrt5\).

Step 3: Add potentials:
\[ V = \frac{1}{4\pi\varepsilon_0}\left[\frac{-q}{r} + \frac{-q}{r} + \frac{q}{r\sqrt5} + \frac{q}{r\sqrt5}\right] = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2q}{r}\left[\frac{1}{\sqrt5} - 1\right] \]

Step 4: Why the other options are wrong.
Options (B) and (D) have \(\frac qr\) with a plus sign inside the bracket, which would make the potential positive, but the nearer charges are negative. Option (C) has \(1 - \sqrt5\), which comes from taking the distances the wrong way round.

Final Answer:
The potential is \(\frac{1}{4\pi\varepsilon_0}\frac{2q}{r}\left[\frac{1}{\sqrt5} - 1\right]\), option (A). \[ \boxed{\frac{1}{4\pi\varepsilon_0}\frac{2q}{r}\left[\frac{1}{\sqrt{5}}-1\right]} \]