Question

A 90 μC point charge is at the origin.

Answer 1

Answer:

a) $E_1=4\times 10^6\ N.C^{-1}$

b) $E_3=202.5\ N.C^{-1}$

Explanation:

Given:

charge at the origin, $q=90\times 10^{-6}\ C$

a)

Electric field at point $(45\ cm,0\ cm)$:

The distance form the charge:

$d_1=0.45\ m$

Now the electric field at the given point:

$E_1=\frac{1}{4\pi.\epsilon_0} \times \frac{q}{d_1^2}$

$E_1=9\times 10^9\times \frac{90\times 10^{-6}}{0.45^2}$

$E_1=4\times 10^6\ N.C^{-1}$

b)

Electric field at point $(-20\ cm, 60\ cm)$:

The distance form the charge:

$d_3=\sqrt{(-20-0)^2+(60-0)^2}$

$d_3=63.2456\ m$

Now the electric field at the given point:

$E_3=\frac{1}{4\pi.\epsilon_0} \times \frac{q}{d_3^2}$

$E_3=9\times 10^9\times \frac{90\times 10^{-6}}{4000}$

$E_3=202.5\ N.C^{-1}$