inner radius 20 cm and A particle with a charge of -60.0 nc is placed at the cen
ID: 1532317 • Letter: I
Question
inner radius 20 cm and A particle with a charge of -60.0 nc is placed at the center of a nonconducting spherical shell of outer radius 21.0 cm. The spherical shell carries charge with a uniform density of -1.99 Ho/m3. A proton moves in a circular orbit just outside the spherical shell. Calculate the speed of the proton. 5.9 10 5 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out a intermediate results to at least four-digit accuracy to minimize roundoff error. m/s Need Help? Read ItExplanation / Answer
(A)
given q = -60 nC , ri = 20.0 cm , ro = 21.0 cm
Field at E1 = q/4*pi*Eo*ro^2
= (60*10^-9)*9*10^9/(0.21)^2
= 12244.89 N/C
Field due to shell is found by intgration ri to ro
E2 = (4/3)*pi*P*(ro^3-ri^3)/4*pi*Eo*ro^2
= (9*10^9)*(4/3*pi*1.99*10^-6)*(0.21^3-0.20^3)/0.21^2
= 2145.16 N/C
total field E = E1 + E2
= 14390.05 N/C
Force on proton = q*E
= 1.6*10^-19*14390.05
= 2.302*10^-15 N
m*v^2/ro = 2.302*10^-15
v = squrt( 2.302*10^-15*0.21)/1.672*10^-27
= 5.37*10^5 m/s
(B)
1. r<a,from Gauss law
the charge enclosed in the loop is the linear charge density
integration (E*ds) = q/E0
E*(2*pi*r*L) = (4*pi)*(lembda*L)/4*pi*E0
E*(2*pi*r*L) = (4*pi*k)*(lembda*L)
E = (2*pi*k)/r
2. a<r<b, the volume of the charge from the cylindrical shell that would be enclosed by the gassian surface is
v = pi*r^2*L -pi*a^2*L
= pi*L(r^2-a^2)
From the Gauss's law
E*(2*pi*r*L) = (4*pi*k)*[(lembda*L)+p*pi*L(r^2-a^2)]
E = 2k*[lembda + p*pi*(r^2-a^2)]/r
3. r>b, the same procedure as above
v = pi*b^2*L -pi*a^2*L
= pi*L(b^2-a^2)
E*(2*pi*r*L) = (4*pi*k)*[(lembda*L)+p*pi*L(b^2-a^2)]
E = 2k*[lembda + p*pi*(b^2-a^2)]/r
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