Charged Sphere Spreadsheet Exercise

Open the Spreadsheet

The Excel spreadsheet named "E-sphere.xls" uses Gauss's law to calculate and graph the strength of the electric field due to a spherical charge distribution. It will calculate the electric field from the center of the spherical distribution out to the distance designated as Rmax; in the spreadsheet, Rmax = 1.0 m. Click here to open the spreadsheet.

Directions

There is a uniform positive charge density of 1.0 . 10-6 C/m3 from R = 0 to R = 0.4 m, and no charge beyond that. The program sums the charge over spherical shells of thickness dR, which is shown to be 0.01 m in cell B3. There are 100 shells. The radii of the surfaces of the shells are listed in cells A10 through A110. In cells B10 through B109, the relative charge density is listed for each shell, next to the value of the radius of the inner surface of the shell. It is called "relative charge density" because the number listed must be multiplied by the value given in cell B5. The spreadsheet does this automatically. The next column, C10 through C110 gives the actual charge contained in each shell, calculated by multiplying the charge density times the volume of the shell. In cells D10 through D110, the total charge contained within the radius given in the corresponding cell in column A is given. This is calculated by adding the values in column C. The electric field at each value of R is listed in cells E10 through E110, calculated with the aid of Gauss's law. Finally, there is a graph of the electric field as a function of R for this charge distribution. A similar graph, together with the derivation of the formulas for the electric field as a function of R is given in sample problem 24-7 on pages 592 and 593 of the textbook.

If the values of the relative charge density in any of the cells B10 through B109 are changed, everything is re-calculated automatically, and the graph is redrawn.

Start by changing the values in cells B10 through B29 to zero. That corresponds to a hollow sphere. In that case, there is no charge from the origin out to 0.20 m, and then a uniform charge density from 0.20 m out to 0.40 m. What do you expect the graph of E vs. R to look like for this distribution? Use Gauss's Law to deduce a formula for the electric field as a function of R for the ranges: 0 < R < 0.20 m, 0.20 m < R < 0.40 m, and R > 0.40 m.

If you fill the empty space from R = 0 to R = 0.20 m with negative charge of the proper relative charge density, it is possible to make the electric field for R > 0.40 m equal to zero. That is, the field due to the negative charge at the center will exactly cancel the field due to the positive outer shell. What relative charge density is required? If you think you know, put the necessary numbers in cells B10 through B29 and see if you are correct. Do not worry about tiny round off errors in the value of the electric field outside of the charged region.

Feel free to try other charge distributions to see what their effects are.

Assignment

The work you turn in should contain:

1. A sketch of the graph of E vs. R for a hollow sphere. No charge from R = 0 to R = 0.2 m, and a uniform charge density from R = 0.2 m to R = 0.4 m.


2. A formula for E as a function of R for the hollow sphere, for each of the ranges: 0 < R < 0.20 m, 0.20 m < R < 0.40 m, and R > 0.40 m.


3. The constant relative charge density for the inner region (R = 0 to 0.2 m) that produces a field of zero for R > 0.4 m


4. A sketch of the graph of E vs. R for a sphere with a negative charge from R = 0 to R = 0.2 m, and positive charge from R = 0.2 m to R = 0.4 m, such that E = 0 for R > 0.4 m.



Now, test yourself with a few practice problems.


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