Question

Particles q1, q2, and q3 are in a straight line.Particles q1 = -5.00 x 10-6 C,92 = +2.50 x 10-6 C,and q3 = -2.50 x 10-6 C. Particles q1 and q2 areseparated by 0.500 m. Particles q2 and q3 areseparated by 0.250 m. What is the net force on 92?Remember: Negative forces (-F) will point LeftPositive forces (+F) will point Right-500 x 100+2.50 x 10°C-2.50 x 10-60q30.500 m0.250 m

Answer 1

Use the following formula for the electric force between two charges:

`F=k\frac{q_{}q^(\prime)_{}}{r^2}`

where

k: Coulomb constant = 9*10^9 Nm^2/C^2

q and q' are charges

r: distance between charges q and q'

The force over q2 is given by the sum of force due to charge 1 and force due to charge 3. Consider that charge q1 and q2 attracts charge q2, then, force between q1 and q2 points to the left and force between q1 and q3 points to the right:

`\begin{gathered} F=-F_(12)+F_(23) \\ F_(12)=(9\cdot10^9(Nm^2)/(C^2))((5\cdot10^(-6)C)(2.50\cdot10^(-6)C))/((0.5m)^2)=0.45N \\ F_(23)=(9\cdot10^9(Nm^2)/(C^2))((2.5\cdot10^(-6)C)(2.50\cdot10^(-6)C))/((0.25m)^2)=0.9N \\ F=-0.45N+0.9N=0.45N \end{gathered}`

Hence, the net force over charge q2 is 0.45N and points to the right.