JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

Two point charges $2 q$ and $q$ are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :

A
$\frac{3 q}{ε _{0}}$
B
$\frac{q}{ε _{0}}$
C
$\frac{3 q}{2 ε _{0}}$
D
$\frac{3 q}{4 ε _{0}}$

Solution

By Gauss's law, $ϕ = \frac{Q _{enclosed}}{ε _{0}}$ , but here both charges sit on the surface, so we use symmetry to find the effective enclosed fraction.

A point charge at a vertex of a cube is shared by 8 cubes, so the effective enclosed share is $\frac{1}{8} (2 q) = \frac{q}{4}$ .

A point charge at the centre of a face is shared by 2 cubes, so the effective enclosed share is $\frac{1}{2} (q) = \frac{q}{2}$ .

Total effective enclosed charge $= \frac{q}{4} + \frac{q}{2} = \frac{3 q}{4}$ .

$∴ ϕ = \frac{3 q}{4 ε _{0}}$

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