(a) Find the charge per unit length (that is, the linear charge density) on the

Question

(a) Find the charge per unit length (that is, the linear charge density) on the inner cylinder.
? nC/m

(b) Calculate the electric field for all values of R.
E 1.5 cm > R > 0 = ? kN/C

E 4.5 cm > R > 1.5 cm = ? (N · m/C )/R

E 6.5 cm > R > 4.5 cm = ? N/C

E r R > 6.5 cm = ? (N · m/C)/R

Suppose that the inner cylinder of the figure above is made of nonconducting material and carries a volume charge distribution given by ?(R) = C/R, where C = 211 nC/m2. The outer cylinder is metallic and both cylinders are infinitely long. (a) Find the charge per unit length (that is, the linear charge density) on the inner cylinder. ? nC/m (b) Calculate the electric field for all values of R. E 1.5 cm > R > 0 = ? kN/C E 4.5 cm > R > 1.5 cm = ? (N m/C )/R E 6.5 cm > R > 4.5 cm = ? N/C E r R > 6.5 cm = ? (N m/C)/R

Explanation / Answer

Picture the Problem We can integrate the density function over the radius of the
inner cylinder to find the charge on it and then calculate the linear charge density
from its definition. To find the electric field for all values of r we can construct a
Gaussian surface in the shape of a cylinder of radius R and length L and apply
Gauss's law to each region of the cable to find the electric field as a function of
the distance from its centerline.

(a) Letting the radius of the
inner cylinder be a, find the
charge Qinner on the inner
cylinder: Qinner = integrate( (R)dV )

Relate this charge to the linear
charge density: = Qinner/L = 2CLa/L
= 2Ca
Substitute numerical values and
evaluate :

= 2(211 nC/m)(0.0150 m)
= 19.9 nC/m


(b) Apply Gauss's law to a
cylindrical surface of radius r and
length L that is concentric with the
infinitely long nonconducting
cylinder: EndA

= 1/ 0 Qinside= 2rLEn = Qinside/ 0

where we've neglected the end areas
because there is no flux through them.
Noting that, due to symmetry,
En = ER,solve for ER to obtain: E R = (Qinside)/ (2RL 0 )

Substitute to obtain, for
R < 1.50 cm: E R < 1.50 cm = (2CLR )/(20LR) = (C )/ 0

Substitute numerical values and
evaluate En(R < 1.50 cm): E R < 1.50 cm = (211 nC/m2)/
(8.854 10-12C2/N · m2)

= 23.8 kN/C



Express Qinside for
1.50 cm < R < 4.50 cm: Qinside = 2CLa

Substitute to obtain, for
1.50 cm < R < 4.50 cm: E 1.50 cm < R < 4.50 cm = (2CaL)/(2 0 RL )
= (Ca )/( 0 R )
where R = 1.50 cm.
Substitute numerical values and evaluate E 1.50 cm < R < 4.50 cm:
E 1.50 cm < R < 4.50 cm = (211 nC/m2)(0.0150 m)
/(8.854 10-12C2/N · m2)r
= (357 N · m/C)/R


Because the outer cylindrical shell
is a conductor: E 4.50 cm < R < 6.50 cm = 0

For R > 6.50 cm, Qinside = 2CLa
and: E R > 6.50 cm = (357 N · m/C)/R

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