Question

A point charge q is at a distance of d/2 directly above the centre of a square of side d . Use Gauss law to obtain the expression for the electric flux through the square . If the point charge is now moved to a distance d from the centre of the square and side of square doubled . Explain how electric flux will be affected .

Answer 1

Here is the answer

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Answer 2

The answers are $\bf \frac{q}{6 \epsilon_o}$ and the electric flux won't change.

Given:

A point charge q at a distance $\frac{d}{2}$ above the centre of a square

A square of side d

To Find:

The electric flux through the square

Solution:

If we assume that there is a cube of side length d, and a charge q is at the centre of the cube, the square will be the bottom face of the cube as the charge q is at a distance $\frac{d}{2}$ above it which is the distance of a face from the centre of a cube.

From Gauss law, we know that the total flux emitting from the charge q is $\frac{q}{\epsilon_o}$.

By symmetry, we know that the flux will equally divide on all the six faces of the cube as the charge is at the centre.

Hence the flux through one face will be $\frac{q}{6 \epsilon_o}$.

If both the distance of the charge and the square length are doubled, we can now assume a cube of side 2d whose centre still has a charge q as the distance of the charge is also doubled. Hence similarly in this case also the flux won't change and remain $\frac{q}{6 \epsilon_o}$.

The electric flux through the square is $\bf \frac{q}{6 \epsilon_o}$ and the electric flux won't change.

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