Question

91 -4.60 x 10-5 C, q2 = +3.10 x 10-6 C, and=93 = -8.30 x 10-5 C. Find the x-component of thenet force on 92. Include the correct + or - sign toindicate direction.0.350 m-91920.155 m93 O- ↓(Make sure you know the direction of each force! Oppositesattract, similar repel.)x-component (N)

Answer 1

We are given the following information.

Charge q1 = -4.60x10-5 C

Charge q2 = +3.10x10-6 C

Charge q3 = -8.30x10-5 C

Distance between q1 and q2 = 0.350 m

Distance between q2 and q3 = 0.155 m

We are asked to find the x-component of the net force on q2.

The net force on q2 is due to the force q1 and q3.

Notice that charge q3 doesn't contribute a force component along the x axis.

So, the net force on q2 only comprises due to force q1.

The force on q2 due to the force q1 is given by

`F=(k\cdot q_1\cdot q_2)/(d_(21)^2)`

Where k is the coulomb's constant (9×10⁹)

`\begin{gathered} F=(9*10^9\cdot-4.60*10^(-5)\cdot3.10*10^(-6))/(0.350^2) \\ F=-10.48\;N \end{gathered}`

The charge q1 contributes an attractive force on the chage q2 (negative x direction).

Therefore, the x-component of the net force on q2 is -10.48 N.