Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 3.3.5

Interactive: Work Along a Path

Home exercise 3.3.5

Is it an electrostatic field?

In spherical coordinates the vector field

E = \frac{A}{r ^{3}} cos θ a_{r} + \frac{B}{r ^{3}} sin θ a_{θ}

is given, where $A$ and $B$ are constants.

(a) Which relation must hold between $A$ and $B$ for $E$ to be an electrostatic field?

(b) Find the corresponding electrostatic potential $V (r, θ)$ , taking $V (\infty) = 0$ .

Hint: an electrostatic field satisfies the first fundamental equation, $\nabla \times E = 0$ , and then $E = - \nabla V$ .

Try it yourself first.

Strategy

Use the test of Section 3.3.3: an electrostatic field must be curl-free. Once the condition on $A$ and $B$ is found, integrate $E = - \nabla V$ component by component (recipe "Is a given field electrostatic?").

An electrostatic field must satisfy the first fundamental equation, $\nabla \times E = 0$ . In spherical coordinates

\nabla \times E = \frac{1}{r ^{2} sin θ} a_{r} \frac{\partial}{\partial r} E_{r} r a_{θ} \frac{\partial}{\partial θ} r E_{θ} r sin θ a_{ϕ} \frac{\partial}{\partial ϕ} r sin θ E_{ϕ}

= \frac{1}{r ^{2} sin θ} a_{r} \frac{\partial}{\partial r} \frac{A}{r ^{3}} cos θ r a_{θ} \frac{\partial}{\partial θ} \frac{B}{r ^{2}} sin θ r sin θ a_{ϕ} \frac{\partial}{\partial ϕ} 0

= \frac{1}{r ^{2} sin θ} r sin θ a_{ϕ} [\frac{\partial}{\partial r} (\frac{B}{r ^{2}} sin θ) - \frac{\partial}{\partial θ} (\frac{A}{r ^{3}} cos θ)] = a_{ϕ} \frac{1}{r} [- \frac{2 B}{r ^{3}} sin θ + \frac{A}{r ^{3}} sin θ] .

The $a_{r}$ and $a_{θ}$ components vanish because $E$ has no $ϕ$ -component and does not depend on $ϕ$ . The remaining component vanishes for all $r$ and $θ$ only if

Answer

A = 2 B

so that

E = \frac{2 B}{r ^{3}} cos θ a_{r} + \frac{B}{r ^{3}} sin θ a_{θ} .

The potential follows from $E = - \nabla V$ , written out in spherical coordinates:

E_{r} a_{r} + E_{θ} a_{θ} + E_{ϕ} a_{ϕ} = - \frac{\partial V}{\partial r} a_{r} - \frac{1}{r} \frac{\partial V}{\partial θ} a_{θ} - \frac{1}{r sin θ} \frac{\partial V}{\partial ϕ} a_{ϕ} .

Comparing components one by one:

\frac{\partial V}{\partial ϕ} = 0 \Rightarrow V = V (r, θ),

\frac{\partial V}{\partial r} = - E_{r} = - \frac{2 B}{r ^{3}} cos θ \Rightarrow V (r, θ) = \frac{B}{r ^{2}} cos θ + f (θ),

\frac{\partial V}{\partial θ} = - \frac{B}{r ^{2}} sin θ + f^{'} (θ) = - r E_{θ} = - \frac{B}{r ^{2}} sin θ \Rightarrow f^{'} (θ) = 0, f (θ) = C .

Then $V (r, θ) = \frac{B}{r ^{2}} cos θ + C$ , and $V (\infty, θ) = 0$ gives $C = 0$ :

Answer

V (r, θ) = \frac{B}{r ^{2}} cos θ [V]

Check

The potential gives back the field:

- \frac{\partial}{\partial r} (\frac{B cos θ}{r ^{2}}) = \frac{2 B cos θ}{r ^{3}} = E_{r}, - \frac{1}{r} \frac{\partial}{\partial θ} (\frac{B cos θ}{r ^{2}}) = \frac{B sin θ}{r ^{3}} = E_{θ} .

Remark

This is the far field of an electric dipole $p = p a_{z}$ with $B = p / (4 π ε_{0})$ (Home exercise 3.1.4, Exercise 3.3.2, Figure 3.3.14b). For $A \neq = 2 B$ the field has a non-zero curl: no potential exists, and the work done around a closed path would not vanish. Such a field cannot be produced by static charges.

For a field with non-zero curl the two paths give different results, and closing the loop leaves a non-zero $\oint E \cdot d l$ , which is exactly what $A \neq = 2 B$ would produce here.

InteractiveLoading simulation…

Home exercise 3.3.5: Is it an electrostatic field?