If Q = 20 $\mu$C and L = 60 cm,what is the magnitude of the electrostatic force on any one of the charges shown?

Question

Quizplus · Accepted Answer

The electrostatic force between two charges is given by Coulomb's law:

F = k * (Q1 * Q2) / r^2

where k is Coulomb's constant, Q1 and Q2 are the charges, and r is the distance between them.

In this case, we have Q = 20 F1F1F1S1μF1F1F10C = 20 * 10^-12 C and L = 60 cm = 0.6 m.

Each charge has a magnitude of Q/10 = 2 * 10^-12 C.

Assuming that the charges are evenly spaced along the rod, the distance between adjacent charges is L/9 = 6.67 cm = 0.0667 m.

Using Coulomb's law and summing the force between each pair of adjacent charges, we get:

F = 8 * k * (Q/10)^2 / (L/9)^2 = 8 * 9 * 10^9 * (2 * 10^-12)^2 / (0.0667)^2

F = 9.1 N

Therefore, the magnitude of the electrostatic force on any one of the charges is 9.1 N.

The best choice is D, which matches the calculation above.

If Q = 20 μ\muμ C and L = 60 Cm,what Is the Magnitude of Cm,what