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JEE Main 2025 (24 Jan, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2025 (24 Jan, Shift 1)

Q32JEE Main 2025 · 24 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

A square loop of sides a=1 ma = 1\text{ m} is held normally in front of a point charge q=1 Cq = 1\text{ C} at a distance a2\frac{a}{2}. The flux of the electric field through the shaded region is 5p×1ε0 Nm2C\frac{5}{p} \times \frac{1}{\varepsilon_0}\,\frac{\text{Nm}^{2}}{\text{C}}, where the value of p is ________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 48

  1. The square is one face of a cube of side aa with qq at its centre: whole-square flux =q6ε0= \frac{q}{6\varepsilon_0}, each quarter =q24ε0= \frac{q}{24\varepsilon_0}.
  2. The shaded region is the lower half (2 quarters) plus one triangle cut by the diagonal of an upper quarter through the centre (half a quarter).
  3. ϕ=2q24ε0+q48ε0=5q48ε0\phi = \frac{2q}{24\varepsilon_0} + \frac{q}{48\varepsilon_0} = \frac{5q}{48\varepsilon_0}, with q=1q = 1 C
  4. So p=48p = 48.
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