Question #295949

A point particle of charge 2.5nC and mass 3.25x10^-3 kg is in a uniform electric field directed to tje right .It is released from rest and moves to the right.After it has traveled 12.0cm,its speed is 25 m/s. Find the

a) work done on the partcle

b) change in tje electric potential energy of the particle and

c) magnitude of the electric field.



1
Expert's answer
2022-02-10T15:05:55-0500

We know that

V2=U2+2aSV^2=U^2+2aS

252=0+2×a×0.1225^2=0+2\times a\times0.12


a=2522×0.12=6252×0.12=2604m/sec2a=\frac{25^2}{2\times0.12}=\frac{625}{2\times0.12}=2604m/sec^2

F=maF=ma


F=3.25×103×2604=8.463NF=3.25\times10^{-3}\times2604=8.463N

Work

W=F.dr

W=8.4263×0.12=1.0155JW=8.4263\times0.12=1.0155J

U=kq1q2rU=-\frac{kq_1q_2}{r}


U=9×109×2.5×2.5×10180.12=4.68×107JU=-\frac{9\times10^9\times2.5\times2.5\times10^{-18}}{0.12}=-4.68\times10^{-7}J

Electric field


E=kqr2=9×109×2.5×1090.122=1562.5N/CE=\frac{kq}{r^2}=\frac{9\times10^9\times2.5\times10^{-9}}{0.12^2}=1562.5N/C


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