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Electrostatics question

2025 · 24 Jan · Shift 2 · Q69
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Electrostatics question

2025 · 24 Jan · Shift 2 · Q69

JEE MainPhysicsElectrostaticsMCQ+4 / −1
JEE Main 2025 (Online) 24th January Evening Shift Physics - Electrostatics Question 22 English In the first configuration (1) as shown in the figure, four identical charges (q0)\left(q_0\right)(q0​) are kept at the corners A, B, C and D of square of side length 'aaa'. In the second configuration (2), the same charges are shifted to mid points G,E,HG, E, HG,E,H and FFF, of the square. If K=14πϵ0K=\frac{1}{4 \pi \epsilon_0}K=4πϵ0​1​, the difference between the potential energies of configuration (2) and (1) is given by :
  1. A
    Kq02a(4−22)\frac{K q_0^2}{a}(4-2 \sqrt{2})aKq02​​(4−22​)
  2. B
    Kq02a(42−2)\frac{K q_0^2}{a}(4 \sqrt{2}-2)aKq02​​(42​−2)
  3. C
    Kq02a(3−2)\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3-\sqrt{2})aKq02​​(3−2​)
  4. D
    Kq02a(32−2)\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)aKq02​​(32​−2)
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