Question

A positive charge Q1=8nC is at the origin, and a second positive charge Q2=12nC is on the x-axis at x=4m. Find:
(i) the net electric field at a point P on the x-axis at x=7m.
(ii) the electric field at a point Q on the y-axis at 3m due to the charges.

Answer 1

i) The electric field is an additive quantity, in our problem it consist of two components:

$$E_1=\frac{kq_1}{(x-x_1)^2},\\ \space\\ E_2=\frac{kq_2}{(x-x_2)^2},\\ \space\\ E=E_1+E_2=\\=k\bigg(\frac{q_1}{(x-x_1)^2}+\frac{q_2}{(x-x_2)^2}\bigg)=\\ \space\\=\frac{1}{4\pi\epsilon_0}\bigg(\frac{8\cdot10^{-9}}{(7-0)^2}+\frac{12\cdot10^{-9}}{(7-4)^2}\bigg)=13\text{ V/m}.$$

ii) We can use the same equations like in i) but make sure you calculate the distance properly:

$$E=k\bigg(\frac{q_1}{(y)^2}+\frac{q_2}{\big(\sqrt{x_2^2+y^2}\big)^2}\bigg)=\\ \space\\=\frac{1}{4\pi\epsilon_0}\bigg(\frac{8\cdot10^{-9}}{(3)^2}+\frac{12\cdot10^{-9}}{(\sqrt{4^2+3^2})^2}\bigg)=12\text{ V/m}.$$