The Magnitude of the Force That One Line Charge Exerts On The Other

[Tachyon]

Hello,

I've been reviewing some general physics and came across a

problem which has stumped me. If there is anyone out ther who can

point me in the right direction, that would be greatly appreciated. The

problem is the following:


Two thin rods of length L lie along the x-axis, one between x = a/2 and
x=a/2 + L, and the other between x = -a/2 and x = -a/2 - L. Each rod has a positive charge Q distributed uniformly along the length. Show that the magnitude of the force that one rod exerts on the other is

F = (Q^2/4*pi*epsilon(0)*L^2)*ln[(a + L)^2/a(a + 2L)]

 

[daniel_i_l]

I think that you just find the forcee as a function of the x position of the rod on the right and integrate from a/2 to L.

 

[kyle8921]

Hello,

To solve this problem, we need to use the formula for the electric force between two point charges, given by Coulomb's Law: F = (k*q1*q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between them. In this problem, we have two rods, each with a positive charge Q distributed uniformly along its length. To find the force between them, we need to divide each rod into tiny point charges, calculate the force between each pair of point charges, and then sum up all these forces to get the total force between the rods.

Let's start by dividing the first rod into tiny point charges. We can do this by dividing the rod into small segments of length dx. The charge on each segment will be dq = (Q/L)*dx, since Q is the total charge on the rod and L is its length. Now, considering a small segment on the first rod at distance x from the origin, the force it exerts on a small segment on the second rod at distance y from the origin will be given by Coulomb's Law as:

dF = (k*dq*dq)/(x-y)^2

To find the total force between the two rods, we need to integrate this expression over all the segments on both rods. The limits of integration for the first rod will be from a/2 to a/2 + L, and for the second rod, it will be from -a/2 to -a/2 - L. So, the total force between the two rods will be given by:

F = ∫∫ dF = ∫∫(k*dq*dq)/(x-y)^2 dx dy

= k*Q^2/L^2 ∫∫(dx*dy)/(x-y)^2

= k*Q^2/L^2 ∫∫(dx*dy)/(x-y)^2

= k*Q^2/L^2 ∫(y-a/2)^2/(y^2-ay+a^2/4)^2 dy

= k*Q^2/L^2 * [ln(y^2-ay+a^2/4) - ln(y-a/2)] |a/2-L, a/2

= k*Q^2/L^2 * [ln((a+L