Consider the reading on coulombs law and on vectors addition. how does the magnitude of the force on ball 1 in this case compare to the force on ball 1 in the previous question?

Force on ball 1 is lesser because the charge is now distributed over the length of the rod. A sketch of the problem can be found in the attachment below. The rod can be assumed to be a continuous distribution of particles that each divide the single charge q2 along the length of the rod. In order to understand this, analysis using the basic definition of columbs law is helpful.

Explanation:

First we divide the rod into infinitesimal segements, each of which has a point charge. Then we drive the expression for the force of interaction of that segement with the charge q1. Then, integrate over the length of the rod from -a to +a.

The detailed step by step solution can be found in the attachment below.

Thank you for reading this and I hope it is helpful to you.

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