Question

A very long conducting pipe (hollow cylinder) has inner radius a= 1.2 cm and outer radius b= 2 cm . It carries charge per unit length of + 530 nC/m . A line of charge lies along the axis of the pipe. The line of charge also has charge per unit length of + 530 nC/m .

A- What is the coefficient of 1/r in the correct expression for the electric field strength as a function of r for r < a?

B- For a < r < b, what is the value of the electric field strength?

C- What is the coefficient of 1/r in the correct expression for the electric field strength as a function of r for r > b?

D- What is the charge per unit length on the inner surface of the tube?

E- What is the charge per unit area on the inner surface of the tube?

F- What is the charge per unit length on the outer surface of the tube?

G- What is the charge per unit area on the outer surface of the tube?

Answer 1

Given that,

inner radius = a = 1.2 cm = 0.012m ; outer radius = b = 2 cm = 0.02 m

charge per unit length = = +530 nC/m = +530 x 10^-9 C/m

A)The electric flux for r < a can be determined by integrating over a closed surface of the pipe. The field will be radial in direction:(lets assume L be the length of the pipe)

= E.dA = E (2 pi r L) = L / o

E = / 2 pi o r = +530 x 10^-9 C/m / 2 x 3.14 x 8.8 x 10^-12 x r = 9590 / r

Hence, the coefficient is 9590 x 1/r.

(b)E for a < r < b The electric field strength is

E = 0

The reason being, field inside any conductor is zero always.

(c)Lets find out the flux again for this case. In this case, the charge per unit length is 2, as the outer and inner got added.

= E.dA = E (2 pi r L) = L(2 )/ o

E = / pi o r = +530 x 10^-9 C/m / 3.14 x 8.8 x 10^-12 r = 19180/r

Hence, coefficeint becomes 19180 x 1/r

(d)We know that the field insie the conductor is always zero. As per the Gauss law the net charge within the gaissian surface should be zero. So to balance at the center there must be equal and oppsite charge density to cancel. Hence, the charge per unit length on the inner surface will be

(in)= - = -530 x 10^-9 C/m = -530 nC/m

(e)We know that

(total) = (in) + (out)

(out) = (total) - (in) = - (-) = 2 = 2 x +530 x 10^-9 = 1060 x 10^-9 C/m = 1060 nC/m

Hence, (out) = 1060 nC/m