Question

In the diagram, 91 = -6.39*10^-9 C andq2 = +3.22*10^-9 C. What is the electricfield at point P? Include a + or - sign tomindicate the direction.q2q1P0.424 m-** 0.636 m(Remember, E points away from + charges,and toward - charges.)(Unit = N/C)Enter

Answer 1

The magnitud of an electric field is given as:

`E=(1)/(4\pi\epsilon_0)(\lvert q\rvert)/(r^2)`

For the charge 1 the magnitude is:

`\begin{gathered} E_1=(1)/(4\pi\epsilon_0)(\lvert-6.39*10^(-9)\rvert)/((0.424)^2) \\ E_1=319.543 \end{gathered}`

Now, since charge 1 is negative this means that this field points towards the charge, in this case to the left, then the electric field for charge one is:

`\vec{E_1}=-319.543`

For the charge 2 the magnitude is:

`\begin{gathered} E_2=(1)/(4\pi\epsilon_0)(\lvert3.22*10^(-9)\rvert)/((0.636)^2) \\ E_2=71.565 \end{gathered}`

Now, since charge 2 is positive this means that this field points away from the charge, in this case to the left, then the electric field for charge two is:

`\vec{E_2}=-71.565`

Now, the total field on point P is the sum of both electric fields, then the total electri field on this point is:

`E=-391.108`