Q132 / 150 · MHT CET 2026 (20 April Shift 1) · Physics

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

  • Correct answer
    A
    $\dfrac{1}{4\pi\epsilon_0}\,\dfrac{2q}{r}\left[\dfrac{1}{\sqrt{5}} - 1\right]$
  • B
    $\dfrac{1}{4\pi\epsilon_0}\,\dfrac{q}{r}\left[\dfrac{1}{\sqrt{5}} + 1\right]$
  • C
    $\dfrac{1}{4\pi\epsilon_0}\,\dfrac{2q}{r}\left[1 - \sqrt{5}\right]$
  • D
    $\dfrac{1}{4\pi\epsilon_0}\,\dfrac{q}{r}\left[1 + \sqrt{5}\right]$

Solution

Let the corners of the square be A, B, C, and D in order, with charges $+q$, $+q$, $-q$, and $-q$ respectively.

The side length of the square is $2r$.

Let P be the midpoint of the side CD, which connects the two negative charges.

The distance of P from the charges at C and D is:

$r_C = r_D = r$

The distance of P from the charges at A and B can be found using the Pythagorean theorem:

$r_A = r_B = \sqrt{r^2 + (2r)^2} = \sqrt{5}r$

The total electric potential at P is the algebraic sum of the potentials due to all four charges:

$V = V_A + V_B + V_C + V_D$

$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q}{\sqrt{5}r} + \dfrac{q}{\sqrt{5}r} - \dfrac{q}{r} - \dfrac{q}{r} \right)$

$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{2q}{\sqrt{5}r} - \dfrac{2q}{r} \right)$

$V = \dfrac{1}{4\pi\epsilon_0} \dfrac{2q}{r} \left[ \dfrac{1}{\sqrt{5}} - 1 \right]$

Answer: $\dfrac{1}{4\pi\epsilon_0}\,\dfrac{2q}{r}\left[\dfrac{1}{\sqrt{5}} - 1\right]$

Hence, the correct answer is option: 'A'.

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