To practice Problem-Solving Strategy 25.2 for continuous charge distribution pro

Question

To practice Problem-Solving Strategy 25.2 for continuous charge distribution problems. A straight rod of length L has a positive charge Q distributed along its length. Find the electric potential due to the rod at a point located a distance d from one end of the rod along the line extending from the rod. (Figure 1) Use superposition to form an algebraic expression for the potential at P. Let the (x, y, z) coordinates remain as variables. Replace the small charge Delta Q with an equivalent expression involving a charge density and a coordinate, such as dx, that describes the shape of charge Delta Q. This is the critical step in making the transition from a sum to an integral because you need a coordinate to serve as an integration variable. Express all distances in terms of the coordinates. Let the sum become an integral. The integration will be over the coordinate variable that is related to Delta Q. The integration limits for this variable will depend on the coordinate system you have chosen. Carry out the integration, and simplify the result. Check that your result is consistent with any limits for which you know what the potential should be. No information is provided on the rod's cross section. For simplicity, assume that its diameter is much smaller than the rod's length You can, then, model the rod as a line of charge. To most efficiently solve this problem, you should divide the rod into pieces of charge that consist of thin lines of charge of length L and a very small cross section. thin "slices" of the rod cut parallel to the axis of the rod. thin "slices" of the rod cut perpendicular to the axis of the rod.

Explanation / Answer

According to the given problem,

Answer:

Thin slices of rod cut perpendicular to the axis of the rod

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