In free space, the retarded potentials are given byV = x(z – ct)V, A = x(z/c – t)a, Wb/mwhere c =aV(a) Prove that V · A = µɛ,at(b) Determine E.

Answered: In free space, the retarded potentials…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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In free space, the retarded potentials are given by
V = x(z – ct)V, A = x(z/c – t)a, Wb/m
where c =
aV
(a) Prove that V · A = µɛ,
at
(b) Determine E.

Expert Solution

Step 1

Given

The given expression for the retarded potentials are

$V = x (z - c t) V A = x (\frac{z}{c} - t) a_{z}$

and $c = \frac{1}{\sqrt{μ_{\circ}} ε_{\circ}}$

Step 2

Now on partially differentiating the voltage equation.

$\begin{array}{rcl} \frac{\partial v}{\partial t} & = & \frac{\partial}{\partial t} (x (z - c t)) \\ = & \frac{\partial}{\partial t} (x z - x c t) \\ = & 0 - x c \\ x & = & - \frac{1}{c} \frac{\partial v}{\partial t} . . . . . . . . . . . . . . . . . . . . . . . . . (i) \end{array}$ {differentiate with respect to t}

Now take the dot product of A with dell operator.

$\begin{array}{rcl} \nabla \cdot A & = & (\frac{\partial}{\partial x} a_{x} + \frac{\partial}{\partial y} a_{y} + \frac{\partial}{\partial z} a_{z}) \cdot (x (\frac{z}{c} - t) a_{z}) \\ = & 0 + 0 + \frac{\partial}{\partial z} (x (\frac{z}{c} - t)) \\ = & \frac{x}{c} \end{array}$ in this case when we multiply different unit vector like $a_{x} \cdot a_{z} = 0$

Now putting value of x from equation(i).

$\begin{array}{rcl} \nabla \cdot A & = & - \frac{1}{c^{2}} \frac{\partial v}{\partial t} \\ = & μ_{\circ} ε_{\circ} \frac{\partial v}{\partial t} \end{array}$ {Hence $c = \frac{1}{\sqrt{μ_{\circ}} ε_{\circ}}$ as given}

Hence $\nabla \cdot A = μ_{\circ} ε_{\circ} \frac{\partial v}{\partial t}$ is proved.

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