1. Force on a Point Charge Due to an Extended Charge Important points to consider:  ‘ ’ is typically used for charge per length  Uniform charge means all little pieces of the charge are the same.  Also, “uniform charge” means:  The limits of integration range from where the charge begins to where it ends. There is no charge to add-up in the space where there is no charge.

2. The Problem: A rod, of length 30 cm, is covered uniformly with 20  C of positive charge. The rod is lying along the + x-axis. The rod’s left end is 10 cm from the origin. A point charge of –5.0  C is located at the origin. Find the force felt by the point charge due to the charged rod. + x + y dq dx d L x

3. Let and because charge is uniform, Now, because the rod lies along the x-axis and has no height or depth, because The expression, describes the force felt between two point charges, so we take a little piece of Q, dq, and treat it as a point charge. Because we are only asking a small piece of the question, dq, we get only a small piece of the answer, dF.

4. Of course, we are usually not satisfied with only a little of the answer so we integrate (sum up) all the little pieces of the answer, dF, to get all of the answer, F. In the spirit of following rules of algebra, if we sum the left side, we must sum the right side: Here, “r” represents the distance from the point charge to dq and because it lies on the x-axis; Now, as we move along adding up our little pieces of dF due to the little (dq)s, we will be taking small steps in the x - direction, dx. That is, “x” is what is changing as be go from one dq to the next dq. Consequently, we must rename dq to dx.

5. Earlier, we determined So,, and now we can rewrite our expression: Pull out the constants: Limits of integration are from the beginning to the end of the rod:

6. Here, the negative sign means repulsion because the charges have unlike signs. This is NOT direction. Direction is determined by your coordinate system.

7. Think about this. We simply substituted values into an equation and got a negative answer. How could that represent direction when this equation is used for all calculations regardless of direction of the rod and whether the point charge or the rod is negative? Our answer would have been the same if the rod had been lying on the negative axis. Again, your coordinate system tells the truth about direction. Incidentally, the electric field, E, points away from the rod. This is the direction a positive charge would move if placed near the rod.