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: Coordinates:
Put the square at $(0,0), (2r,0), (2r,2r), (0,2r)$ with the negative charges at $(0,0)$ and $(2r,0)$. The midpoint is $P = (r, 0)$.

Step 2: Distances from P:
To $(0,0)$: $r$. To $(2r,0)$: $r$. To $(2r,2r)$: $\sqrt{r^2 + 4r^2} = r\sqrt5$. To $(0,2r)$: $r\sqrt5$.

Step 3: Sum:
$V = k\left[-\frac{2q}{r} + \frac{2q}{r\sqrt5}\right] = \frac{k\,2q}{r}\left(\frac{1}{\sqrt5} - 1\right)$, where $k = \frac{1}{4\pi\varepsilon_0}$. This is negative, as expected since the negative charges are closer.

Final Answer:
Option (A) gives the potential. \[ \boxed{\frac{1}{4\pi\varepsilon_0}\frac{2q}{r}\left[\frac{1}{\sqrt{5}}-1\right]} \]