An infinite line charge of uniform density l = 4 µC/m runs along the symmetry ax
ID: 1665841 • Letter: A
Question
An infinite line charge of uniform density l = 4 µC/m runs along the symmetry axis(z-axis) of a thin, infinitely long, non-conductingcylindrical shell of radius a1 = 0.1 m. Thelatter carries a uniform charge density s1 = 3 µC/ m2.Surrounding the latter and concentric with it is a thick,infinitely long, conducting cylindrical shell with inner radiusa 2 = 0.7 m and outer radiusa3 = 0.8 m. The conducting cylinder isuncharged.
(a) Calculate the charge per unit length on the non-conductingshell.
l1 = C/m
(b) Calculate the electric field at the following values ofr, where r is the radial distance measuredperpendicularly from the symmetry axis.
At r = 0.05 m:
E = N/C
At r = 0.4 m:
E = N/C
At r = 0.75 m:
E = N/C
At r = 1.6 m:
E = N/C
(c) Calculate the surface charge densities on the inner andouter surfaces of the conducting shell.
Inner: s2 = C/m2
Outer: s3 = C/m2
Explanation / Answer
a)The surface charge density s1=3C/m^2 the charges per unit length of its iss1*2a1=1,88C/m=1,88e-6C/m=1 b)1.r=0,05 m According to Gauss'law E*2r=/0 E=/(02r)=1,44e6(V/m) 2.r=0,4 m, this point is in between the two shells. According to Gauss'law E*2r=(+1)/0=2,64e5(V/m) 3.r=0,75m, this point is inside the conducting shell, so there isno electric field E=0 4.r=1,6m According to Gauss'law E*2r=(+1)/0=6,61e4(V/m) c).Since the electric field inside the outer shell is zero. so the total charge in that Gaussian'surface is zero. so the charge per unit length on the inner surface is-(1+)=-5,88C=2 the charge density 2=2/(2a2)=-1,34e-6(C/m2) d), similar to above, 3=-2(2a3)=1,17e-6(C/m2)