- #1

yungman

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This is from Field and Waves Electromagnetic by Cheng page 602. THis is regarding to elemental dipoles. Given: [Quote Book]

[tex] i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7)}[/tex]

Since the current vanishes at the ends of the wire, charge must be deposited there.

[tex] i(t)=^+_-\frac {d q(t)}{dt} \;\;\hbox { (11.8)}[/tex]

In phasor notation, [itex]q(t)=\Re e[Qe^{j\omega t}][/itex], we have

[tex]I=^+_-j\omega Q\;\;\hbox{ (11.9)}[/tex]

or [tex]Q = ^+ _- \frac {I} {j \omega}\;\;\hbox {(11.10)}[/tex]

[End Book]

This is what my calculation of q(t):

[tex] i(t)=\frac {d [q(t)]}{dt} = \Re e \left [ Q \frac {d (e^{j\omega t})}{dt}\right ] =\Re e [ Q j\omega e^{j\omega t}] = -\omega \;Q \;sin\;\omega t [/tex]

You see [itex] i(t)[/itex] not equal to [itex]I\; cos\;\omega t [/itex]

I just don't see how this work! Please help.

[tex] i(t)=\Re e [Ie^{j\omega t}]=I\; cos\;\omega t\;\;\hbox { (11.7)}[/tex]

Since the current vanishes at the ends of the wire, charge must be deposited there.

[tex] i(t)=^+_-\frac {d q(t)}{dt} \;\;\hbox { (11.8)}[/tex]

In phasor notation, [itex]q(t)=\Re e[Qe^{j\omega t}][/itex], we have

[tex]I=^+_-j\omega Q\;\;\hbox{ (11.9)}[/tex]

or [tex]Q = ^+ _- \frac {I} {j \omega}\;\;\hbox {(11.10)}[/tex]

[End Book]

This is what my calculation of q(t):

[tex] i(t)=\frac {d [q(t)]}{dt} = \Re e \left [ Q \frac {d (e^{j\omega t})}{dt}\right ] =\Re e [ Q j\omega e^{j\omega t}] = -\omega \;Q \;sin\;\omega t [/tex]

You see [itex] i(t)[/itex] not equal to [itex]I\; cos\;\omega t [/itex]

I just don't see how this work! Please help.

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