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*10^(-6)\text{ C} \end{gathered}`

The other charge is,

`\begin{gathered} q_2=-38.06\text{ }\mu C \\ =-38.06*10^(-6)\text{ C} \end{gathered}`

The second charge is at,

`x=1.04\text{ m}`

To find:

The electric field at,

`x=(1.04)/(3)\text{ m}`

Step-by-step 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=(kq_1)/(d^2) \\ Here,\text{ k=9}*10^9\text{ N.m}^2.C^(-2) \\ d=(1.04)/(3)\text{ m} \end{gathered}`

The electric field due to the first charge is,

`\begin{gathered} E_1=(9*10^9*11.63*10^(-6))/(((1.04)/(3))^2) \\ =870.9*10^3\text{ N/C along the first charge} \end{gathered}`

The electric field due to the second charge is,

`\begin{gathered} E_2=(9*10^9*38.06*10^(-6))/((1.04-(1.04)/(3))^2) \\ =712.57*10^3\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)*10^3 \\ =158.33*10^3\text{ N/C} \end{gathered}`

Hence, the magnitude of the electric field is

`158.33*10^3\text{ N/C}`