Purpose: We will study the potential from a continuous charge distribution and find how to calculate it different ways. In the end each calculate will produce the same answer. The method we will use are integration and approximation by summing up several finite elements of charge q by using excel. Finally, use Gauss's law to find the electric field along with the defining equation for potential difference to set up a line integral.
We started by defining an equation for the charge distribution on a ring on a point. The ring has radius a and the distance from the ring to the point is x. We found that V=sqrt(x^2 + a^2)
We were given that the ring had 20 charges on it so we divided them equal through the whole ring.
We started by calculating the potential difference by hand (pictured below), we then used excel to calculate the same problem, and finally we derived an integral which we used to calculate the same problem. All methods yield the same solution.
| Here we have the our solution to the electric potential by hand. |
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| Here is our excel solution |
| Here we derived the electric potential of the ring on a point distance x, which we found to be equal to kqcos(theta) divided by x. |
| Our final solution was that E is equal to kQx divided by (x^2 + a^2)^3/2 |
In the next part of class, we begin our study of the electric potential on a charged rod. We are given an observation point at (0.1, 0.15). Picture below is our solution.
Equipotential Surfaces
Sometimes it is possible to move along a surface without doing any work. Thus, it is possible to remain at the same potential energy anywhere along such a surface. If an electric charge can travel along a surface without doing any work, the surface is called an equipotential surface.




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