Question

Two charges are separated by 1.04 m. One charge is -11.63 μC located at x=0. The other charge is -38.06 μC located at 1.04 m. What is the magnitude of the electric field at x=1.04/3 m?

Answer 1

Given:

The charge at x=0 is,

$\begin{gathered} q_1=-11.63\text{ }\mu C \\ =-11.63\times10^{-6}\text{ C} \end{gathered}$

The other charge is,

$\begin{gathered} q_2=-38.06\text{ }\mu C \\ =-38.06\times10^{-6}\text{ C} \end{gathered}$

The second charge is at,

$x=1.04\text{ m}$

To find:

The electric field at,

$x=\frac{1.04}{3}\text{ m}$

Explanation:

The diagram of the charges is shown below:

The electric field at the given point due to the first charge is,

$\begin{gathered} E_1=\frac{kq_1}{d^2} \\ Here,\text{ k=9}\times10^9\text{ N.m}^2.C^{-2} \\ d=\frac{1.04}{3}\text{ m} \end{gathered}$

The electric field due to the first charge is,

$\begin{gathered} E_1=\frac{9\times10^9\times11.63\times10^{-6}}{(\frac{1.04}{3})^2} \\ =870.9\times10^3\text{ N/C along the first charge} \end{gathered}$

The electric field due to the second charge is,

$\begin{gathered} E_2=\frac{9\times10^9\times38.06\times10^{-6}}{(1.04-\frac{1.04}{3})^2} \\ =712.57\times10^3\text{ N/C along the second charge} \end{gathered}$

The electric fields are opposite each other. So, the net electric field is,

$\begin{gathered} E_1-E_2 \\ =(870.9-712.57)\times10^3 \\ =158.33\times10^3\text{ N/C} \end{gathered}$

Hence, the magnitude of the electric field is

$158.33\times10^3\text{ N/C}$