Question

Two fixed charges, A and B, are located at x axis. A is at (-3 m, 0 m), B is at x = (+3 m, 0 m). QA = +4.0 μC and QB = -4.0 μC. Calculate the electric field at point (0 m, 4 m) on the y axis.

Answer 1

Answer:
`2.88* 10^3`

Step-by-step explanation:

Given

Charge on A is
`4 \mu C` at (-3,0)

Charge on B is
`-4 \mu C` at (3,0)

Electric Field at a distance d is given by
`E=(kQ)/(d^2)`

Thus Electric filed at (0,4) due to A is

`E_A=(kQ_A)/(d^2)=(1)/(4\pi \epsilon )(Q_A)/(d^2)`

Here
`d =√(3^2+4^2)=5`

`E_A=(9* 10^9* 4* 10^(-6))/(5^2)`

`E_A=1.44* 10^3 N\C (Away\ From\ Q_A)`

Similarly
`E_B` is
`1.44* 10^3 N\C (towards Q_B)`

Electric Field is at an angle
`\theta`which is given by

`tan\theta =(4)/(3)`

Thus
`E_Asin\theta`and
`E_Bcos\theta`component will cancel out each other and Cos component will add up

`E_(net)=E_Acos\theta +E_Bcos\theta`

`E_(net)=2Ecos\theta`

`E_(net)=2* 1.44* 10^3 N\C` (towards positive x axis )