JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 26)
A point charge of +12$$\mu$$C is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be __________ $$\times$$ 103 Nm2/C.
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Answer
226
Explanation
Given, charge, q = 12 $$\mu$$C = 12 $$\times$$ 10$$-$$6C
Height of charge from surface, h = 6 cm = 6 $$\times$$ 10$$-$$2 m and side of square, a = 12 cm = 12 $$\times$$ 10$$-$$2 m
Using Gauss law, it is a part of cube of side 12 cm and charge at centre so;
$$\phi = {Q \over {6{\varepsilon _0}}} = {{12\mu c} \over {6{\varepsilon _0}}} = 2 \times 4\pi \times 9 \times {10^9} \times {10^{ - 6}}$$
$$ = 226 \times {10^3}$$ Nm2 / C
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