Question

Three objects, carrying charges of 4.0 x 10^6 C, 6.0 x 10^6 C, and +9.0 x 10^6 C, respectively, are placed in a line, equally spaced from left to right by a distance of 0.50m. Calculate the magnitude and direction of the net force acting on each charge that results from the presence of the other two

Answer 1

`F_1=540*10^(-12) N`

`F_2=-1080 * 10^(-12) N`

`F_3=-2268 * 10^(-12) N`

Step-by-step explanation:

From the question we are told that

Charge on objects

`Q_1=4.0 * 10^6 C`

`Q_2= 6.0 * 10^6 C`

`Q_3= +9.0 * 10^6 C`

Distance apart
`d=0.50m`

Generally the equation for Force on individual Charge is mathematically given by

For Q_1

`F_1=(d*Q_1Q_2)/(r^2)+( d*Q_1Q_3)/(4r^2)`

`F_1=(k*(4.0 * 10^6 C)(6.0 * 10^6 C))/(0.5^2)+( d*(4.0 * 10^6 C)( +9.0 * 10^6 C))/(4(0.5)^2)`

`F_1=540*10^(-12)`

For Q_2

`F_2= (k)/(0.5^2[Q_1 + Q_3]Q_2)`

`F_2= (k)/(0.5^2[4.0 * 10^6 C + 9.0 * 10^6 ]*6.0 * 10^6 C)`

`F_2=-1080 * 10^(-12) N`

For Q_3

`F_3=(kQ_3)/(r^2[Q_1/4+Q_2)`

`F_3=(k(+9.0 x 10^6 C))/((0.5)^2[(4.0 x 10^6 )/4+(6.0 x 10^6 C))`

`F_3=-2268 * 10^(-12) N`