Question

Wo thin uniformly charged rods, each with length L and total charge +Q, are parallel and separated by a distance a. The first rod has one end at the origin and its other end on the positive y-axis. The second rod has its lower end on the positive x-axis

Determine the x-component of the differential force dF2 exerted on a small portion of the second rod, with length dy2 and position y2, by the first rod. (This requires integrating over differential portions of the first rod, parameterized by dy1. )

Answer 1

To find the x-component of the differential force exerted on a small portion of the second rod by the first rod, we can integrate over differential portions of the first rod using Coulomb's law.

Step-by-step explanation:

To determine the x-component of the differential force dF2 exerted on a small portion of the second rod, we need to integrate over differential portions of the first rod. Let's consider a small element of the charge distribution between y and y + dy. The charge in this cell is dq = λ dy and the distance from the cell to the field point P is √(x² + y²).

To find the x-component of the differential force dF2, we can use Coulomb's law:

dF2 = k * (dq1 * dq2) / (r²)

By integrating over the length of the first rod, with limits of integration from 0 to L, we can find the net force on the second rod.