Potential of Continuous Charge Distributions (4/16/2015)

Potential Difference Due to Continuous Charge Distributions

To start, we have a ring where Q = 20 µC and the radius is a = 30 cm. These numbers come into play later, because first we were asked to formulate an equation which would solve for the potential difference V at point P, where x = 20 cm. In the first image, our calculations to solve for this equation are shown. The final correct answer is actually written in black on the side, where V = kDq/square-root(x^2+a^2). The purpose here is to see one of the ways to solve for finding potential difference, where we can use an integral method in which the potential dV from each element of charge dq is integrated mathematically to give a total potential at the location P. But first we will discretely calculate V in Excel.

Estimate the Potential from a Charged Ring

Shown here is the second method of solving for the same ring. We can approximate the value of the potential V by summing up several finite elements of charge Dq by using a computer spreadsheet. We divided the ring into 20 elements of charge Dq and calculated the total V at a distance of x = 20 cm from the center. The final answer we found was V = 4.99*10^5.

Calculation of the Potential from a Charged Ring

To further analyze how to approach this problem, we were tasked with figuring out the equations needed to solve for V at points P, when P is at the top of the ring, the middle, and the bottom. The correct equations are shown in the picture on the right.

What we realized when solving for P at the top of the ring, was that we ended up with an angle to solve for. Depicted above-right is the proper equation for the angle. Two other equations for the angle are possible, written as theta = arccos(x/sqrt(x^2+4a^2)) and theta = arcsin(2a/sqrt(x^2+4a^2)).

The final step in this long process, after calculating all possible equations and variables to be used, is to finally calculate for the actual potential difference V and compare it to what we found through the spreadsheet. The final answer is shown above, but without rounding it comes to 4.99*10^5 V, which is the exact same value as the one presented on Excel.

DV from a Ring Using the E-Field Method (Gauss' Law)

The third, and final, way to find potential difference of the charged ring is to use Gauss' Law in a way that will allow us to solve for V. Shown above is the derivation for converting Gauss' Law into a form that applies to potential difference. 

(r=x^2+a^2 should be r^2=x^2+a^2)

After plugging in the appropriate values, we again end up with the correct value for potential difference. We now have three different approaches which we can use to solve for the same unknown.

Electric Potential Due to a Finite-Length Line-Charge

For the next activity, we are tasked with finding the total charge and potential difference of a 16.0 cm long, horizontal rod has a uniform charge density of 2.70 mC/cm, while taking the left end of the rod to be the origin. We also need to determine the magnitude of the potential at (10cm, 15cm).

We have calculated the total charge discretely using Excel as shown above. By separating the rod into 16 segments, and integrating through the length of the bar from 0.055cm to 0.095cm, we can calculate the charge at each segment and add them together for the total charge of 2.46*10^6 V. This is the value of electric potential at any point on the bar.

In order to solve for the magnitude of potential at (10cm,15cm), the above equation for V needs to be used, which takes into account the distance r from the origin of the bar to the point. The equation for V with variables for a and b was actually derived using Wolfram Alpha, and afterwards we simply plugged in our appropriate values. Not displayed above is the answer for the potential, which is 2.43*10^4 V.

Sketches of Electrical Field Lines and Equipotentials

If an electric charge can travel along a surface without doing any work, the surface is called an equipotential surface. Pictured above are three equipotential surfaces depicted as the line going through the center of the first drawing (going clockwise from left, to right), around the center of the second drawing, and along the curvature of the third drawing. The premise is that the path of the charge and the electric field lines are perpendicular of one another, resulting in the equation E(dot)ds= 0.

Electric Potential Lab/Activity

Power Source

Voltmeter

Conductive Paper

Pictured above are the various pieces of equipment used to run this activity. We had a power source connected to two points on the conductive paper, which we then turned on to run a voltage through. We then used a voltmeter to measure the magnitude of the voltage at various distances. This was done by placing both of the voltmeter needles are precise lengths from each other. The Position values shown below depict the distance and direction that one of the needles moved in relation to the other stationary needle, which was left at the origin (one of the metal circles on the paper). It should be noted that the potential difference between the two metal circles is 15.08 V. 15 V was the amount of power we chose to produce from the power supply.

After gathering this data, we were asked to answer several questions in the lab notebook shown below:

Summary:

In summary, we have established three different methods of finding potential difference V between two points: discretely using a spreadsheet, algebraically/trigonometrically by solving for V, and solving for V using Gauss' Law. We have also looked into what it means for a surface to be equipotential and how a charge can do zero work by traveling along such a surface. We also got some hands on experience by manually testing potentially differences between two points on conductive paper. This allowed us to gather some data and calculate how much work a charge would have to do to move from one point on the paper to another using the equation W=QDV.