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FIGURE P25.69 shows a thin rod of length L and charge Q. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod.

Short Answer

Expert verified

The electric potential is given asV=kQLIn2x+L2x-L

Step by step solution

01

Given Information

We need to find an expression for the electric potential a distancex away from the center of the rod on the axis of the rod.

02

Simplify

Consider the fact that the potential is a scalar quantity, which means

V=kqr

We can express this as an integral. In our case, instead of the charge, we'll have the linear charge density Q/L, and the integration variable, let it bel. The potential at point xwill therefore be given as

V=kQ/Ll+x-L2L2dl

Solving this integral,

V=kQLdll+x-L/2L/2=kQLlnl+x-L/2L/2=kQLlnx+L/2-ln(xL/2

By the properties of the logarithm and by simplifying with the factor of two, we get

V=kQLIn2x+L2x-L

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