Experiment 3. Please derive all equations and show how to calculate uncertainty.

Question

Experiment 3. Please derive all equations and show how to calculate uncertainty. Data from experiment 2 is as followed.
Mass: 0.0001kg Distance between the two cylinders: 0.0367m Length of string: 0.125m Angle of string: 8.4 degrees Charge on cylinder:( 4.66 x 10^-9 Coulombs ) Lab #2 instructions are also included for reference Due Date Experiment 3 Electric Force, Electric Field, Potential Energy and Potential Difference Qbiectives . To use your measured data from Experiment 2 to determine the olectric forces electric fields, potential energies, and electric potentials generated by the charged cylinders Introduction This "experiment is essentially a continuation of Experiment 2. No new measurements are required. Again, you should clearly state any assumptions and/or approximations that you use in your derivation. All calculations should include uncertainties. Electric Fonce Derive an equation in terms of known and measured quantities for the force magnitude that each cylinder exerts on the other. Then, use this equation to calculate this force magnitude. Electric Field Derive an equation for the electric field along a vertical (y) axis passing . [Be equidistant between the two cylinders in terms of m, d, and L (and y, of course) sure that your equation gives you the expected results at y o, yeoo, and ysod.] Use your equation to find an equation for electric field when y d/2. Calculate the field magnitude at this point using your data. Electrical Potential Eneray Derive an equation for electrical potential energy in terms of m, d, and L. Then, calculate the electrical potential energy of this two charge configuration Derive an equation for electric potential at the midpoint between the charges in erms of m, d, and L. Then, calculate this electric potential.

Explanation / Answer

Given mass m = 0.0001 kg
distance between two cylinde3rs, d = 0.0367 m
length of string, l = 0.125 m
Angle, theta = 8.4 deg
charge, q = 4.66*10^-9 C

so, force on any one cylinder would be F = kq^2/d^2

1. calculating F = 8.98*10^9*(4.66*10^-9)^2/0.0367^2 = 144.7824 *10^-6 N
2. along the vertical axis, the two cylinders will cancel out each others horizontal component of force and pnly vertical compoennts will be left which will add up
   so, Ey = 2*ykq/(d^2/4 + y^2)^3/2
   also,
   F/mg = tan(theta)
   kq^2/d^2 = F
   kq^2/d^2*mg = d/2*sqroot(l^2 - d^2/4)

   hence
   q^2 = d^3*mg/2k*sqroot(l^2 - d^2/4)

   so, Ey = 2y*k*sqroot([ d^3*mg/2k*sqroot(l^2 - d^2/4)])/(d^2/4 + y^2)^3/2

3. electric potential energy = kq^2/d = d^2*mg/2*sqroot(l^2 - d^2/4)
   PE = 5.343 micro J

4. electric potential at midpoint = 4kq/d = 4*k*sqroot(d^3*mg/2k*sqroot(l^2 - d^2/4))/d

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