An infinitely long, cylindrical, insulating shell of inner radius a and outer ra
ID: 1328737 • Letter: A
Question
An infinitely long, cylindrical, insulating shell of inner radius a and outer radius b has a uniform volume charge density sigma. A lone of uniform linear charge density lambda is placed along the axis of the shell. Determine the electric field in the following regions. (Choose a gaussian cylinder of radius r and length L. Use any variable or symbol stated above with the following as necessary: ke) r LT a magnitude e 2klambda/r direction a LT r LT b magnitude e=25(lambda+ppi(r2+a(2)))/r direction outward r GT b magnitude E=2k(lambda+ppi(b2+a(2)))/r direction outwordExplanation / Answer
Infinitely long cylindrical insulating shell,
inner radius = a , outer radius = b.
Consider the Gaussian surface with lenth L, uniform volume charge density = row and linear charge density = lamda.
Since Electric feild is normal to the surface of cylindrical shell,
case 1) r < a , charge enclosed by this region is line charge lamda,
By Gauss Law, intergral of E . da = (integral of lamda . dl) / eps0
E . 4 pi r2 = lamda L / eps0
E = lamda L / ( 4 pi eps0 r2)
E = ke lamda L / r2 where ke = 1 / 4pi eps0
Electric field pointing outward from centre.
case 2) a < r < b , charge density row over the volume
integral of E . da = (integral of arrow . dv - integral of lamda . dl) / eps0
E . 4pi r2 = (row pi L (r2 - a2) - lamda L) / eps0
E = ( row pi L (r2 - a2) - lamda L) / 4 pi eps0 r2
E = ke ( row pi L (r2 - a2) - lamda L) / r2
Elecrtric field is pointing outwaed from centre.
case 3) r > b, line charge over the outer surface
integral of E. da = (integral of lamda . dl + integral of ab row . dv) / eps0
E . 4pi r2 = ( lamda L + row pi L (b2 - a2 ) / eps0
E = ( lamda L + row pi L (b2 - a2) / 4pi eps0 r2
E =ke ( lamda L + row pi L(b2 - a2 ) / r2
Electric field is pointing out ward from centre.