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Two thin uniformly charged rods, each with length Land 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 external work required to move the second rod from very far away to the position x=a, provided the first rod is held fixed at x = 0.

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Final answer:

The external work required to move a uniformly charged rod in an electric field can be found by calculating the electric potential due to the other rod and then integrating this potential across the length of the moving rod to find the total work done.

Step-by-step explanation:

The question pertains to the concept of electric potential energy in the field of physics. To determine the external work required to move the second charged rod from infinity to a final position x=a, we can use the concept that work done is equal to the change in potential energy of the system when moving a charge in an electric field.

First, one must calculate the electric potential V produced by the first rod at a distance a from it. Then, the total work W required can be calculated by integrating the infinitesimal work done dW to move an infinitesimal charge dq from the second rod over its entire length L.

The expression for work done by moving a charge in an electric field created by a continuous charge distribution is W = λ∫Vdx, where λ is the charge per unit length on the second rod and V is the electric potential due to the first rod.

For a uniformly charged rod, the electric potential V at a point along the perpendicular bisector at distance a from the rod depends on the total charge Q, the length of the rod L, and fundamental constants. The work done can thus be found by integrating the potential V across the length of the second charged rod. This requires the usage of calculus, specifically integration, due to the continuous charge distribution.

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