Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

em 24 have hen will be de them on ting? al plates is e separated ge on each ii r

ID: 1441076 • Letter: E

Question

em 24 have hen will be de them on ting? al plates is e separated ge on each ii r and r: an and 01. ectric field I charge e he noncon then slowly n a time t (a) just me ample 22-4 he sphere's distributed c is also a the electric Problem 33 34. (I) A very long solid nonconducting cylinder of radius R and length (Ro possesses a uniform volume charge density ph (C/m), Fg. 22-34. Dctermine the electric field at points (a) outside the cylinder (R Ro) and (l) inside the cylinder (R Ro). Do only for points far from the ends and for which R FIGURE 22-34 Problem 34 35. A thin cylindrical shell of radius Ri is surrounded by a sccond concentric cylindrical shcll of radius R2 (Fig 22-350 The inner shell has a total charge +Q and the outer shell -e. Assuming the length t of the shells is much greater than R1 or R2, determine the electric field as a function of R (the perpendicular distance from the commo n axis of the cylin R., and (c) R: R2 ders) for (a) 0

Explanation / Answer

35.

a) considering a cylinder about same axis of radius R and length L as gaussian surface.

now applying Gauss law,

flux = E.A = Qinside / e0

for 0 < R < R1

Qinside = 0   (there is no charge inside R1 radius)

hence flux = E.A = 0

E = 0

b) R1 < R2 < R2

Qin = (Q/t) L

E ( 2pi R L ) = (QL/t) / e0

E = Q / (2 pi e0 t R)

c) for R > R2

Qin = Q + (-Q) = 0

E = 0

d) for R = (R1 + R2) / 2

R1 < R < R2

hence E = Q / (2 pi e0 t R) = Q / (pi e0 t (R1 + R2))

force on electron, F = qE

and this force wil work as centripetal force

F = m v^2 / R

Q / (2 pi e0 t R) = m v^2 / R


m v^2 = Q / (2 pi e0 t)

and KE = m v^2 / 2 = Q / (4 pi e0 t)