1 pasted at the origin. You have to apply an inward force to push the q. ( r) q :

Transcription

1 letromagneti Theory (MT) Prof Ruiz, UNC Asheville, dotorphys on YouTube Chapter U Notes nergy U nergy in an letri Field ring a harge q to a distane r away from q Consider the two harges positive and q pasted at the origin You have to apply an inward fore to push the q in sine q r r r F applied dl q dl + q V ( r) q V ( r) V ( ) these harges repel So you have to push against the outward fore e take the potential at infinity as our zero referene q V ( r) The potential V ( r ) is due to q 4π r q : V ( r) To emphasize that the distane r is between the two harges we write V ( r) q 4π r The work is q q 4π r Now we bring in a third harge to the two harges q and q already in plae extra q3 Vnew ( r), where extra qq3 qq 3 + 4π r3 r3 The total work so far is 4π q q q q q q r r r Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

2 The general formula for n harges is 4 4 n n i π i ri i 4 q q n n i π i ri > i q q q n n q i i π ri i n qv i ( r i ) i ( r ) V ( r ) d ρ ρ ( ) V ( r ) d V ( ) d ( fa) A f + f A f A ( fa) A f V ( V) V ( V) d Vd Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

3 ( V) d ( ) d ( V) d + d F da Fd Use the Divergene Theorem on the first term V da + d Take the volume to be large as any volume will do as long you enlose the harges ut the potential is zero at infinity So for the super large volume the first integral vanishes Here is what is left d all spae all spae d nergy per unit volume is given the the next equation u PU (Pratie Problem) From this equation, ustify that the energy stored in a parallelplate apaitor is Ad U nergy in a Magneti Field Faraday's Law in Integral Form: dφ dl L di Power for the Current: P IV To fight the bak MF so we an produe the urrent, we apply Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

4 d di giving applied Papplied I LI applied LI Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense LI Faraday's Law in Differential Form (a Maxwell quation): Magneti Field from Vetor Potential: ( A) t A t A t t A ut don't forget the basi prodution of eletri fields from stati harges: V So in general A V t A t For eletri fields generated by magneti flux hanges we ust need dφ dl an be expressed as dφ di L leads to Summary: d d A dl di A dl L and A dl LI LI and A dl LI

5 I A dl I J da J nda ˆ The urrent diretion is normal to the ross setion: I A nˆ dl J nˆ da A nˆ dl J Adadl J Ad or µ J A ( ) d µ dl l A Jd ˆn ( A ) eˆ ( ˆ n ik Ai ek ) δ nk ( ik Ai ) ( ik Ai ) x x x n n k A i ( A ) ( ik Ai ) ik + ik Ai x x x k k k Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

6 Ai ( A ) ki + ki Ai xk Ai ( A ) ki ki Ai x k x k x ( A ) ( A) A ( ) ( A ) ( A) A ( ) A ( ) d µ A ( ) ( A) ( A ) A ( ) ( A ) d ( A ) d µ µ d ( A ) da µ µ Large volume gives dereasing vetor potential and magneti fields k d µ µ d nergy per unit volume: u µ PU (Pratie Problem) From this equation, ustify that the energy stored in a oil with urrent is µ Al Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

7 U3 nergy in an letromagneti ave Summary from an earlier Chapter t t µ λ f sin [ k ( z t ) ] i sin k( z t) k π λ M ave Courtesy Pwormer, ikimedia t t t sin [ k ( z t ) ] i Faraday's Law in Differential Form (a Maxwell quation): iˆ ˆ kˆ ˆ k os k( x t) x y z sin k( z t) k os k( x t) ˆ Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense

8 k ˆ os k( x t) and t ˆ k os k( x t) sin [ k( z t) ] t t k os k( x t) ˆ os k( z t) ( k) k ( ) k µ sin [ k ( z t ) ] i sin [ k( z t) ] nergy per unit volume: u + µ µ u and u µ u µ µ Then, u k z t S µ sin ( ) sin [ k( z t) ] kˆ Define S ukˆ Poynting vetor ("points" in wave diretion) PU3 (Pratie Problem) Show that the unit for the Poynting vetor is energy per unit area per unit time Mihael J Ruiz, Creative Commons Attribution-NonCommerial-ShareAlike 3 Unported Liense