I am comparing potential energy to kenetic energy using 1/2 CV^2 = 1/2 mv^2, whe

Question

I am comparing potential energy to kenetic energy using 1/2 CV^2 = 1/2 mv^2, where C is the capacitance of a capacitor, V is the voltage stored in the capacitor, m is the mass of an object, and v is the velocity of the oblect. How could I find the error and what type of error should I use if I had the following values: C = 9000 microfarads and I know there is error on that but iam not sure of it or how to find it, I am using a CBS-90-010, V = 20.21 +/- .01 volts, m = .0001 +/- .00001 kg, and v = 6.99 +/- .08 m/s

Explanation / Answer

Here we have 1/2 CV^2 = 1/2 mv^2

    or C= mv^2/V^2

       ln C = ln [mv^2/V^2] = ln m + 2ln v - 2ln V

taking derivative of both sides,

    dC/C = dm/m + 2 dv/v + 2dV/V because errors add up, dont cancel.

dc/9000 = 0.00001/0.0001 + 2*0.08/6.99 + 2*0.01/20.21

        dC = [0.00001/0.0001 + 2*0.08/6.99 + 2*0.01/20.21]*9000

             = 1115 microfarad Answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.