Question

A point charge Q1=3uc at origin and a point charge Q2=-7 uc is placed at x= 0.4m on the x axis or cartesian system.determine the electric field E(x)=E(x)x at all points along x axis

Answer 1

Let's draw the rough picture of the charges lies on the X axis,

Clearly, electric field due to charge $Q_1=3 \mu C$ at $R$ is

$$E_1=\frac{1}{4 \pi \epsilon_0}\frac{Q_1}{(x+0.4)^2} \hspace{1cm}(1)$$

similarly, electric field due to charge $Q_2=-7 \mu C$ at $R$ is

$$E_2= \frac{1}{4 \pi \epsilon_0}\frac{Q_2}{x^2} \hspace{1cm}(2)$$

Now, by principle of superposition total electric field at $R$ will be

$$E_{total}=\frac{1}{4 \pi \epsilon_0}\bigg(\frac{Q_1}{(x+0.4)^2}+\frac{Q_2}{x^2} \bigg)\\ \implies E_{total}=\frac{1}{4 \pi \epsilon_0}\bigg(\frac{3\times 10^{-6}}{(x+0.4)^2}+\frac{-7 \times 10^{-6}}{x^2} \bigg)N/C\\ \implies E_{total}=\frac{1\times10^{-6}}{4 \pi \epsilon_0}\bigg(\frac{3}{(x+0.4)^2}-\frac{7 }{x^2} \bigg) N/C\\$$

Direction of $E_{total}$ is along the negative $x$ axis.

Hence,we are done.