Question

Two surface OABC and OCDE lies in the plane of xy and yz as shown in the figure. A charged particle 'q' lies in the space at a point P. if

• A. coordinates of 'P' is (a − Δr, a − Δr, Δr) and a >> Δr, then flux passing through surface OABC is $\frac{7q}{24\epsilon_0}$
• B. coordinates of 'P' is (a − Δr, a − Δr, Δr) and a >> Δr, then flux passing through surface OCDE is $\frac{q}{24\epsilon_0}$
• C. coordinates of 'P' is (a + Δr, a + Δr, −Δr) and a >> Δr, then flux passing through surface OABC is $\frac{q}{24\epsilon_0}$
• D. coordinates of 'P' is (a + Δr, a + Δr, −Δr) and a >> Δr, then flux passing through surface OCDE is $\frac{q}{24\epsilon_0}$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The problem involves calculating the electric flux through two surfaces, OABC and OCDE, due to a point charge 'q' at point P. We use symmetry arguments for a point charge placed near a corner of a cube.

Key Concepts:

• Gauss's Law: The total electric flux through a closed surface is proportional to the charge enclosed.
• Symmetry: Exploiting symmetry simplifies flux calculations.
• Charge near a Corner: A charge near a corner of a cube has specific flux distributions through the faces.

Case A and B: P(a - Δr, a - Δr, Δr) and a >> Δr (Charge inside the cube near a corner)

This places the charge 'q' infinitesimally inside the cube near corner B(a,a,0). In this scenario:

• Flux through OABC: $\frac{7q}{24\epsilon_0}$
• Flux through OCDE: $\frac{q}{24\epsilon_0}$

Case C and D: P(a + Δr, a + Δr, -Δr) and a >> Δr (Charge outside the cube near a corner)

This places the charge 'q' infinitesimally outside the cube near corner B(a,a,0). In this scenario:

• Flux through OABC: $\frac{q}{24\epsilon_0}$
• Flux through OCDE: $\frac{-q}{24\epsilon_0}$

Therefore, options A, B, and C are correct.