5 Engineering Example 8

5.1 Field strength on a cylinder

Problem in words

Given the electric field E ̲ on the surface of a cylinder, use Gauss’ law to find the charge per unit length.

Mathematical statement of problem

On the surface of a long cylinder of radius a , the electric field is given by

E ̲ = ρ L 2 π ε 0 a + b cos θ r ̂ ̲ − b sin θ θ ̲ ( a 2 + 2 a b cos θ + b 2 )

(using cylindrical polar co-ordinates) due to a line of charge a distance b ( < a ) from the centre of the cylinder. Using Gauss’ law , find the charge per unit length.

Find the integral ∫ ∫ E ̲ ⋅ d S ̲ and by equating this to Q ε 0 , find b in the expression for E ̲ , using the result

∫ 0 2 π a + b cos θ ( a 2 + 2 a b cos θ + b 2 ) d θ = 2 π a

Mathematical analysis

Consider a cylindrical section - as in the previous example, there are no contributions from the ends of the cylinder since the electric field has no normal component here. However, on the curved surface

d S ̲ = a d θ d z r ̂ ̲

So

E ̲ ⋅ d S ̲ = ρ L 2 π ε 0 a + b cos θ ( a 2 + 2 a b cos θ + b 2 ) a d θ d z

Integrating over the curved surface of the cylinder

∫ ∫ S E ̲ ⋅ d S ̲ = ∫ z = 0 l ∫ θ = 0 θ = 2 π a ρ L 2 π ε 0 a + b cos θ ( a 2 + 2 a b cos θ + b 2 ) d θ d z = a ρ L l 2 π ε 0 ∫ 0 2 π a + b cos θ ( a 2 + 2 a b cos θ + b 2 ) d θ = ρ L l ε 0  using the given result.

From Gauss’ law

ρ L l ε 0 = Q ε 0 so Q = ρ L l

Interpretation

Therefore the charge per unit length on the line of charge is given by ρ L (i.e. the charge per unit length is constant).

Task!

Verify Gauss’ theorem for the vector field F ̲ = x i ̲ − y j ̲ + z k ̲ and the unit cube 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤ z ≤ 1 .

  1. Find the vector ∇ ̲ ⋅ F ̲ .
  2. Evaluate the integral ∫ z = 0 1 ∫ y = 0 1 ∫ x = 0 1 ∇ ̲ ⋅ F ̲ d x d y d z .
  3. For each side, evaluate the normal vector d S ̲ and the surface integral ∫ ∫ S F ̲ ⋅ d S ̲ .
  4. Show that the two sides of the statement of the theorem are equal.
  1. 1
  2. 1
  3. − d x d y k ̲ , 0 ; d x d y k ̲ , 1 ; − d x d y j ̲ , 0 ; d x d z j ̲ , − 1 ; − d y d z i ̲ , 0 ; d y d z i ̲ , 1
  4. Both sides are 1 .
Exercises
  1. Verify Gauss’ theorem for the vector field F ̲ = 4 x z i ̲ − y 2 j ̲ + y z k ̲ and the cuboid 0 ≤ x ≤ 2 , 0 ≤ y ≤ 3 , 0 ≤ z ≤ 4 .
  2. Verify Gauss’ theorem, using cylindrical polar coordinates, for the vector field F ̲ = ρ − 2 ρ ̂ ̲ over the cylinder 0 ≤ ρ ≤ r 0 , − 1 ≤ z ≤ 1 for
    1. r 0 = 1
    2. r 0 = 2
  3. For S being the surface of the tetrahedron with vertices at ( 0 , 0 , 0 ) , ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) and ( 0 , 0 , 1 ) , find the surface integral

    ∫ ∫ S ( x i ̲ + y z j ̲ ) ⋅ d S ̲
    1. directly
    2. by using Gauss’ theorem

    Hint :- When evaluating directly, show that the unit normal on the sloping face is 1 3 ( i ̲ + j ̲ + k ̲ ) and that d S ̲ = ( i ̲ + j ̲ + k ̲ ) d x d y

  1. Both sides are 156 ,
  2. Both sides equal (a)  4 π ,  (b)  2 π ,
  3. Both sides equal 5 24 .