JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Electrostatics

A point charge of

+ 12 μ C

is at a distance 6 cm vertically above the centre of a squared of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be...........

\times 1 0^{3} Nm^{2} / C

Answer: 226

Solution

The charge is

6

cm above the centre of a square of side

12

cm, which places it exactly at the centre of a cube of edge

12

cm having that square as one face. By symmetry the total flux

\frac{q}{ε _{0}}

divides equally among the six faces, so

Φ = \frac{1}{6} \cdot \frac{q}{ε _{0}} = \frac{12 \times 1 0 ^{- 6}}{6 \times 8.85 \times 1 0 ^{- 12}} = 2.26 \times 1 0^{5} N m^{2} C^{- 1}

The magnitude of the flux through the square is

226 \times 1 0^{3} N m^{2} /C

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer

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