An infinitely long, cylindrica insulating shell of inner radius a and outer radi

Question

An infinitely long, cylindrica insulating shell of inner radius a and outer radius b has a uniform volume charge density p. A line of uniform linear charge density is placed Determine the electric field in the following regions (Choose a gaussian cylinder of the she radius r and length L. se any able or symbol stated along the axis of above along with the following as necessary: ke.) (a) r a 3 2 magnitude E Outward direction (b) a r b magnitude E direction no direction x (c) r b magnitude E e direction Select Need Help? Read It

Explanation / Answer

a) from Gauss Law,

flux = E.A = q_inside / e0

for r < a

q_inside = lambda * L

so E ( 2pi r L ) = (lambda*L) / e0

E = (lambda) / 2 pi e0 r

and ke = 1/4pie0 and lambda = linear charge density

E = 2 ke lambda / r

charge is +ve so outward. ........ANS

B) a < r < b

q_inside = (lambda * L ) + (pL * (pi (r^2 - a^2) / (pi (b^2 - a^2))

= (lambda*L + pL(r^2 - a^2)/(b^2 - a^2) )

E ( 2pi r L ) = (lambda*L + pL(r^2 - a^2)/(b^2 - a^2) ) / e0

E = (lambda + p(r^2 - a^2)/(b^2 - a^2) ) / 2 pi e0 r

E = 2 ke ( lambda(b^2 -a^2) + p(r^2 -a^2) ) / (b^2 - a^2)

outward

c) r > b

q_inside = (lambda*L + p pi (b^2 - a^2 )L )

E ( 2pi r L ) = (lambda*L + p pi (b^2 - a^2 )L ) / e0

E = (lambda / 2pi e0 r) + ( p pi (b^2 - a^2) / 2pi e0 r )

E = (2 ke lambda / r ) + ( 2 k3 pi p (b^2 - a^2) / r )

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---