Question

Two point charges are fixed on the y axis a negative point charge q1-26 μC at y1 = +0.21 m and a positive point charge q2 art y2 = +0.39 m. A third point charge q = +8.1 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 28 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

`3.13* 10^(-5) C`

Step-by-step explanation:

We are given that

`q_1=26\mu C=26* 10^(-6) C`

`1\mu C=10^(-6) C`

`y_1=0.21 m`

`q=8.1\mu C=8.1* 10^(-6) C`

`y_2=0.39 m`

F=28 N

We have to find the magnitude of q2.

We know that

`F=(kq_1q_2)/(r^2)`

Where
`k=9* 10^9`

Using the formula

Force on charge q due to charge q1

`F_1=(9* 10^9* 26* 10^(-6)* 8.1* 10^(-6))/((0.21)^2)`

`F_1=42.98 N`

Force on charge q due to point charge q2

`F_2=(9* 10^9* q_2* 8.1* 10^(-6))/((0.39)^2)`

`F_2=4.79* 10^5 q_2`

`F=F_1-F_2`

`28=42.98-4.79* 10^5q_2`

`4.79* 10^5q_2=42.98-28=14.98`

`q_2=(14.98)/(4.79* 10^5)=3.13* 10^(-5) C`