A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is: (a) (q)/(24ε₀) (b) (q)/(48ε₀) (c) (q)/(4ε₀) (d) (q)/(8ε₀)

Answer: (a). A charge at a corner is shared by 8 cubes; the three far faces each carry 1/3·(q)/(8ε₀) = (q)/(24ε₀). The shaded top triangle is cut by the diagonal through the charge's projection, so it is exactly half of a face: (q)/(48ε₀). Taking the side triangle also as half a face: (q)/(48ε₀). Total = (q)/(48ε₀) + (q)/(48ε₀) = (q)/(24ε₀), answer (a).

A charge 'q' is placed at one corner of a cube as shown in figure… | JEE Main 2021 (25 Feb, Shift 2)

Q85JEE Main 2021 · 25 FEB · SHIFT 2Medium

Electric Flux & Gauss's Law

A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\vec{E}$ through the shaded area is:

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Answer: (a)

A charge at a corner is shared by 8 cubes; the three far faces each carry $\frac{1}{3}\cdot\frac{q}{8\varepsilon_0} = \frac{q}{24\varepsilon_0}$ .
The shaded top triangle is cut by the diagonal through the charge's projection, so it is exactly half of a face: $\frac{q}{48\varepsilon_0}$ .
Taking the side triangle also as half a face: $\frac{q}{48\varepsilon_0}$ .
Total $= \frac{q}{48\varepsilon_0} + \frac{q}{48\varepsilon_0} = \frac{q}{24\varepsilon_0}$ , answer (a).

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