Question

Two point charges of 60.0 C and -12.0 C are separated by a distance of 20.0 cm. A 7.00 C charge is placed midway between these two charges. What is the electric force acting on this charge because of the other two charges

Answer 1

`4.5* 10^(14) N`

Step-by-step explanation:

`q_1=60 C`

`q_2=-12 C`

`q_3=7 C`

Distance between q1 and q2,d=20 cm

`r=(d)/(2)=(20)/(2)=10 cm=10* 10^(-2) m`

1m =100 cm

Electric force on charge q3 due to charge q1

`F_1=(kq_1q_3)/(r^2)=(9* 10^9* 60* 7)/((10* 10^(-2))^2)` (away from q1)

Electric force on charge q3 due to charge q2

`F_2=(kq_2q_3)/(r^2)=(9* 10^9* 12* 7)/((10* 10^(-2))^2)`(attract toward q2)

Net force=
`F=F_1+F_2`

`F=(9* 10^9* 60* 7)/((10* 10^(-2))^2)+(9* 10^9* 12* 7)/((10* 10^(-2))^2)`

`F=(9* 10^9* 7)/((10* 10^(-2))^2)(60+12)`

`F=4.5* 10^(14) N`