A student collected the following data on the period ofosccillation of a charged

Question

A student collected the following data on the period ofosccillation of a charged particle Assume mass of the particle is m = 1.23 10-27kg, a= 0.01 cm. Calculate an average value of period. Place your finalanswer for Average period in a box. (c) Manipulate the above equation so that it is solvedfor  q. (d) Plug in the numbers using your average period found inpart (b) (show them plugged into your formula) and find a value ofq. Place your final answer for q in a box. Calculate a percenterror with the accepted value of 1.6 x 10-19 C.

Explanation / Answer

(b)The average value of period is T = (T1 + T2 + T3/3)------------------(1) Here,T1= 1.41 * 10-5 s,T2=1.03 * 10-5 s and T3= 1.10 * 10-5s Substituting the above values in equation (1),we get T = (1.41 * 10-5 + 1.03 * 10-5 + 1.10 *10-5/3) or T = 1.18 * 10-5 s (c)The period of oscillation of the charged particle is givenby T = 2 * (m * a3/2K * q2) or q2= 2 * (m * a3/2K * T) or q = [2 * (m * a3/2K * T)]1/2-------------------(2) Here,m = 1.23 * 10-27 kg,a = 0.01 cm = 0.01 *10-2 m,K = (1/4o) = 9 *109 Nm2/C2 and T = 1.18 *10-5 s (d)Substituting the above values in equation (2),we get q = [2 * 3.14 * (1.23 * 10-27 * (0.01 *10-2)3/2 * 9 * 109 * 1.18 *10-5)]1/2 or q = 0.190 * 10-21 C = 0.00190 * 10-19C or q = 0.00190 * 10-19 C The accepted value of q is 1.6 * 10-19 C The percent error with the accepted value of q is (0.00190 * 10-19/1.6 * 10-19) *100 or 0.12 % or q = 0.190 * 10-21 C = 0.00190 * 10-19C or q = 0.00190 * 10-19 C The accepted value of q is 1.6 * 10-19 C The percent error with the accepted value of q is (0.00190 * 10-19/1.6 * 10-19) *100 or 0.12 %

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