Question

hello experts

problem- A charge Q is placed at the centre of a square if electric field intensity due to the charge at the corners of the square is 'E' and the intensity at the mid point of the side of square is 'e'
then ratio of E/e will be?
NOTE- the Electric field mentioned here are different at both the points.

thanking you

Answer 1

Answer on Question#53115 - Physics - Field theory

A charge $Q$ is placed at the centre of a square if electric field intensity due to the charge at the corners of the square is 'E' and the intensity at the mid-point of the side of square is 'e' then ratio of $E/e$ will be?

NOTE- the Electric field mentioned here are different at both the points.

Solution:

Let the side of the square be $a$, then the distance from the centre of the square to its corner is $d_c = \frac{a}{\sqrt{2}}$ and the distance from the centre to the mid-point of the side is $d_m = \frac{a}{2}$. Therefore, the electric field intensity due to the charge at the corner of the square is given by

$$E = \frac{Q}{4\pi\varepsilon_0 d_c^2}$$

and at the mid-point of the side

$$e = \frac{Q}{4\pi\varepsilon_0 d_m^2}$$

So

$$\frac{E}{e} = \frac{d_m^2}{d_c^2} = \frac{\left(\frac{a}{2}\right)^2}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{1}{2}$$

Answer: 1/2.