6 Engineering Example 3

6.1 Electric potential

Introduction

There is a scalar quantity V , called the electric potential, which satisfies

̲ V = E ̲ where E ̲ is the electric field.

It is often easier to handle scalar fields rather than vector fields. It is therefore convenient to work with V and then derive E ̲ from it.

Problem in words

Given the electric potential, find the electric field.

Mathematical statement of problem

For a point charge, Q , the potential V is given by

V = Q 4 π ϵ 0 r

Verify, using spherical polar coordinates, that E ̲ = ̲ V is indeed E ̲ = Q 4 π ϵ 0 r 2 r ̂ ̲

Mathematical analysis

In spherical polar coordinates:

̲ V = V r r ̂ ̲ + 1 r V θ θ ̂ ̲ + 1 r sin ϕ V ϕ ϕ ̂ ̲ = V r r ̂ ̲  as the other partial derivatives are zero = r Q 4 π ϵ 0 r r ̂ ̲ = Q 4 π ϵ 0 r 2 r ̂ ̲

Interpretation

So E ̲ = Q 4 π ϵ 0 r 2 r ̂ ̲ as required.

This is a form of Coulomb’s Law. A positive charge will experience a positive repulsion radially outwards in the field of another positive charge.

Example 26

Using spherical polar coordinates, find ̲ F ̲ for the following vector functions.

  1. F ̲ = r r ̂ ̲
  2. F ̲ = r 2 sin θ r ̂ ̲
  3. F ̲ = r sin θ r ̂ ̲ + r 2 sin ϕ θ ̂ ̲ + r cos θ ϕ ̂ ̲
Solution
  1. ̲ F ̲ = 1 r 2 sin θ r ( r 2 sin θ F r ) + θ ( r sin θ F θ ) + ϕ ( r F ϕ ) = 1 r 2 sin θ r ( r 2 sin θ × r ) + θ ( r sin θ × 0 ) + ϕ ( r × 0 ) = 1 r 2 sin θ r ( r 3 sin θ ) + θ ( 0 ) + ϕ ( 0 ) = 1 r 2 sin θ 3 r 2 sin θ + 0 + 0 = 3

    Note :- in Cartesian coordinates, the corresponding vector is F ̲ = x i ̲ + y j ̲ + z k ̲ with ̲ F ̲ = 1 + 1 + 1 = 3 (hence consistency).

  2. ̲ F ̲ = 1 r 2 sin θ r ( r 2 sin θ F r ) + θ ( r sin θ F θ ) + ϕ ( r F ϕ ) = 1 r 2 sin θ r ( r 2 sin θ r 2 sin θ ) + θ ( r sin θ × 0 ) + ϕ ( r × 0 ) = 1 r 2 sin θ r ( r 4 sin 2 θ ) + θ ( 0 ) + ϕ ( 0 ) = 1 r 2 sin θ 4 r 3 sin 2 θ + 0 + 0 = 4 r sin θ
  3. ̲ F ̲ = 1 r 2 sin θ r ( r 2 sin θ F r ) + θ ( r sin θ F θ ) + ϕ ( r F ϕ ) = 1 r 2 sin θ r ( r 2 sin θ r sin θ ) + θ ( r sin θ × r 2 sin ϕ ) + ϕ ( r × r cos θ ) = 1 r 2 sin θ r ( r 3 sin 2 θ ) + θ ( r 3 sin θ sin ϕ ) + ϕ ( r 2 cos ϕ ) = 1 r 2 sin θ 3 r 2 sin 2 θ + r 3 cos θ sin ϕ + 0 = 3 sin θ + r cot θ sin ϕ
Example 27

Find ̲ × F ̲ for the following vector fields F ̲ .

  1. F ̲ = r k r ̂ ̲ , where k is a constant
  2. F ̲ = r 2 cos θ r ̂ ̲ + sin θ θ ̂ ̲ + sin 2 θ ϕ ̂ ̲
Solution
  1. ̲ × F ̲ = 1 r 2 sin θ r ̂ ̲ r θ ̂ ̲ r 2 sin θ ϕ ̂ ̲ r θ ϕ F r r F θ r 2 sin θ F ϕ = 1 r 2 sin θ r ̂ ̲ r θ ̂ ̲ r 2 sin θ ϕ ̂ ̲ r θ ϕ r k r × 0 r 2 sin θ × 0 = 1 r 2 sin θ θ ( 0 ) ϕ ( 0 ) r ̂ ̲ + ϕ ( r k ) r ( 0 ) r θ ̂ ̲ + r ( 0 ) θ ( r k ) r 2 sin θ ϕ ̂ ̲ = 0 r ̂ ̲ + 0 θ ̂ ̲ + 0 ϕ ̂ ̲ = 0 ̲
  2. ̲ × F ̲ = 1 r 2 sin θ r ̂ ̲ r θ ̂ ̲ r sin θ ϕ ̂ ̲ r θ ϕ F r r F θ r sin θ F ϕ = 1 r 2 sin θ r ̂ ̲ r θ ̂ ̲ r sin θ ϕ ̂ ̲ r θ ϕ r 2 cos θ r × sin θ r sin θ × sin 2 θ = 1 r 2 sin θ θ ( r sin 3 θ ) ϕ ( r sin θ ) r ̂ ̲ + ϕ ( r 2 cos θ ) r ( r sin 3 θ ) r θ ̂ ̲ + r ( r sin θ ) θ ( r 2 cos θ ) r sin θ ϕ ̂ ̲ = 1 r 2 sin θ 3 r sin 2 θ cos θ + 0 r ̂ ̲ + 0 sin 3 θ r θ ̂ ̲ + sin θ + r sin θ r sin θ ϕ ̂ ̲ = 3 sin θ cos θ r r ̂ ̲ sin 2 θ r θ ̂ ̲ + ( 1 + r ) r sin θ ϕ ̂ ̲
Task!

Using spherical polar coordinates, find ̲ f for

  1. f = r 4
  2. f = r r 2 + 1
  3. f = r 2 sin 2 θ cos ϕ
  1. 4 r 3 r ̂ ̲ ,
  2. 1 r 2 ( 1 + r 2 ) 2 r ̂ ̲ ,
  3. r ( r 2 sin 2 θ cos ϕ ) r ̂ ̲ + 1 r θ ( r 2 sin 2 θ cos ϕ ) ϕ ̂ ̲ + 1 r 2 sin θ ϕ ( r 2 sin 2 θ cos ϕ )

    = 2 r sin 2 θ cos ϕ r ̂ ̲ + 2 r cos 2 θ cos ϕ θ ̂ ̲ 2 r cos θ sin ϕ ϕ ̂ ̲

Exercises
  1. For F ̲ = r sin θ r ̂ ̲ + r cos ϕ θ ̂ ̲ + r sin ϕ ϕ ̂ ̲ , find ̲ F ̲ and ̲ × F ̲ .
  2. For F ̲ = r 4 cos θ r ̂ ̲ + r 4 sin θ θ ̂ ̲ , find ̲ F ̲ and ̲ × F ̲ .
  3. For F ̲ = r 2 cos θ r ̂ ̲ + cos ϕ θ ̂ ̲ find ̲ ( ̲ × F ̲ ) .
  1. cos ϕ ( cot θ + c o s e c θ ) + 3 sin θ , cot θ 2 sin ϕ r ̂ ̲ 2 sin ϕ θ ̂ ̲ + ( 2 cos ϕ cos θ ) ϕ ̂ ̲
  2. 0 , 2 r 5 sin θ ϕ ̂ ̲
  3. 0