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Question: A point charge \(+ 20\mu C\) is at a distance 6 cm directly above the centre of a square of side 12 ...

A point charge +20μC+ 20\mu C is at a distance 6 cm directly above the centre of a square of side 12 cm as shown is figure. The magnitude of electric flux through the square is.

A

2.5×106Nm2C−12.5 \times 10^{6}Nm^{2}C^{- 1}

B

3.8×105Nm2C−13.8 \times 10^{5}Nm^{2}C^{- 1}

C

4.2×105Nm2C−14.2 \times 10^{5}Nm^{2}C^{- 1}

D

2.9×106Nm2C−12.9 \times 10^{6}Nm^{2}C^{- 1}

Answer

3.8×105Nm2C−13.8 \times 10^{5}Nm^{2}C^{- 1}

Explanation

Solution

: Here, take the square as one face of the cube with edge 12 cm. the point charge +20 μ\muC is at a distance 6 cm directly above the centre of ABCD From figure, it is clear that square ABCD is one of the six faces of a cube of side 12 cm. By Gauss’s theorem, total electric flux therough all the six gaces of the cube

=qε0= \frac{q}{\varepsilon_{0}}

∴\therefore Electric flux through the square

φ=16qε0\varphi = \frac{1}{6}\frac{q}{\varepsilon_{0}}

= 16×2.×10−68.85×10−12\frac{1}{6} \times \frac{2. \times 10^{- 6}}{8.85 \times 10^{- 12}}

=3.8×105Nm2C−1= 3.8 \times 10^{5}Nm^{2}C^{- 1}