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

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Coordinates of ‘P’ is $(a - Δ r, a - Δ r, Δ r)$ and $a > > Δ r$ , then flux passing through surface OABC is $\frac{7 q}{24 ε_{0}}$ .

Coordinates of ‘P’ is $(a - Δ r, a - Δ r, Δ r)$ and $a > > Δ r$ , then flux passing through surface OCDE is $\frac{q}{24 ε_{0}}$ .

Coordinates of ‘P’ is $(a + Δ r, a + Δ r, - Δ r)$ and $a > > Δ r$ , then flux passing through surface OABC is $\frac{q}{24 ε_{0}}$ .

Coordinates of ‘P’ is $(a + Δ r, a + Δ r, - Δ r)$ and $a > > Δ r$ , then flux passing through surface OCDE is $\frac{q}{24 ε_{0}}$ .

answer is A, B, C, D.

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Detailed Solution

Using symmetry if charged particle lies at $P_{1}$ then $ϕ_{O C D E} = \frac{q}{24 ε_{0}}$
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and $ϕ_{O A B C} = \frac{(\frac{q}{ε_{0}} - \frac{q}{8 ε_{0}})}{3} = \frac{7 q}{24 ε_{0}}$
If the charge particle lies at $P_{2}$ then $ϕ_{O C D E} = \frac{q}{24 ε_{0}}$ and $ϕ_{O A B C} = \frac{q}{24 ε_{0}}$