Question

Three point charges are on the x axis: q1 = -6.0 µC is at x = -3.0 m, q2 = 1.0 µC is at the origin, and q3 = -1.0 µC is at x = 3.0 m. Find the electric force on q1.

Answer 1

The electric force on q₁ is
`4.5*10^(-3)\ N`.

Step-by-step explanation:

Given that,

Charge on
`q_(1)=-6.0\ \mu C`

Distance
`x= -3.0\ m`

Charge on
`q_(2)=1.0\ \mu C` at origin

Distance
`x= 3.0\ m`

Charge on
`q_(3)=-1.0\ \mu C`

We need to calculate the electric force on q₁

Using formula of electric force

`F_(12)=(kq_(1)q_(2))/(r^2)`

Put the value into the formula

`F_(12)=(9*10^(9)*(-6.0*10^(-6))*1.0*10^(-6))/((3.0)^2)`

`F_(12)=-0.006\ N`

Negative sign shows the attraction force.

We need to calculate the electric force F₁₃

`F_(13)=(9*10^(9)*(-6.0*10^(-6))*(-1.0*10^(-6)))/((6.0)^2)`

`F_(13)=0.0015\ N`

Positive sign shows the repulsive force.

We need to calculate the net electric force

`F=F_(12)+F_(13)`

`F=-0.006+0.0015`

`F=0.0045\ N`

`F=4.5*10^(-3)\ N`

Hence, The electric force on q₁ is
`4.5*10^(-3)\ N`.