Question

A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between the two charges. Use k = 9 x 10^9 N*m^2/C^2.

Answer 1

Step-by-step explanation:

Given

Charge of first Particle
`q_1=+12\ \mu C`

Charge of second Particle
`q_2=-8\ \mu C`

distance between them
`d=4\ cm`

`k=9* 10^(9)`

magnetic field due to first charge at mid-way between two charged particles is

`E_1=(kq_1)/(r^2)`

`r=(d)/(2)=(4)/(2)=2\ cm`

`E_1=(9* 10^9* 12* 10^(-6))/((2* 10^(-2))^2)`

`E_1=27* 10^7\ N/C` (away from it)

Electric field due to
`q_2=-8\ \mu C`

`E_2=(kq_2)/(r^2)`

`E_2=-(9* 10^9* 8* 10^(-6))/((2* 10^(-2))^2)`

`E_2=-18* 10^7\ N/C`(towards it)

`E_(net)=E_1+E_2`

`E_(net)=9* 10^7\ N/C`(away from first charge)