Electric Potential. Electric Potential Video: Section 1 4. Electric Fields and WORK 9/3/2014. IB Physics SL (Year Two) Wednesday, September 3, 2014


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1 9/3/014 lectric Potentil IB Physics SL (Yer Two) Wenesy, Septemer 3, 014 lectric Potentil Vieo: Section 1 4 lectric Fiels n WORK In orer to rin two like chres ner ech other work must e one. In orer to seprte two opposite chres, work must e one. Rememer tht whenever work ets one, enery chnes form. As the monkey oes work on the positive chre, he increses the enery of tht chre. The closer he rins it, the more electricl potentil enery it hs. When he releses the chre, work ets one on the chre which chnes its enery from electricl potentil enery to kinetic enery. very time he rins the chre ck, he oes work on the chre. If he rouht the chre closer to the other oject, it woul hve more electricl potentil enery. If he rouht or 3 chres inste of one, then he woul hve h to o more work so he woul hve crete more electricl potentil enery. lectricl potentil enery coul e mesure in Joules just like ny other form of enery. 1
2 9/3/014 lectric Fiels n WORK Consier netive chre movin in etween oppositely chre prllel pltes initil K=0 Finl K= 0, therefore in this cse Work = DP We cll this LCTRICAL potentil enery,, n it is equl to the mount of work one y the LCTRIC FORC, cuse y the LCTRIC FILD over istnce,, which in this cse is the plte seprtion istnce. Is there symolic reltionship with the FORMLA for rvittionl potentil enery? lectric Potentil mh ( or W ) m q h x ( W ) q W q Here we see the eqution for rvittionl potentil enery. Inste of rvittionl potentil enery we re tlkin out LCTRIC POTNTIAL NRGY A chre will e in the fiel inste of mss The fiel will e n LCTRIC FILD inste of rvittionl fiel The isplcement is the sme in ny reference frme n use vrious symols Puttin it ll toether! Question: Wht oes the LFT sie of the eqution men in wors? The mount of nery per chre! nery per chre The mount of enery per chre hs specific nme n it is clle, VOLTAG or LCTRIC POTNTIAL (ifference). Why the ifference? 1 W DK mv q q q
3 9/3/014 nerstnin Difference Let s sy we hve proton plce etween set of chre pltes. If the proton is hel fixe t the positive plte, the LCTRIC FILD will pply FORC on the proton (chre). Since like chres repel, the proton is consiere to hve hih potentil (volte) similr to ein ove the roun. It moves towrs the netive plte or low potentil (volte). The pltes re chre usin ttery source where one sie is positive n the other is netive. The positive sie is t 9V, for exmple, n the netive sie is t 0V. So siclly the chre trvels throuh chne in volte much like fllin mss experiences chne in heiht. (Note: The electron oes the opposite) BWAR!!!!!! W is lectric Potentil nery (Joules) is not V is lectric Potentil (Joules/Coulom).k. Volte, Potentil Difference The other sie of tht eqution? mh ( or W ) m q h x ( W ) q W q Since the mount of enery per chre is clle lectric Potentil, or Volte, the prouct of the electric fiel n isplcement is lso VOLTAG This mkes sense s it is pplie usully to set of PARALLL PLATS. = 3
4 9/3/014 xmple A pir of oppositely chre, prllel pltes re seprte y 5.33 mm. A potentil ifference of 600 V exists etween the pltes. () Wht is the mnitue of the electric fiel strenth etween the pltes? () Wht is the mnitue of the force on n electron etween the pltes? m 600V? 19 q 1.6x10 C e Fe Fe 600 ( ) 19 q 1.6x10 C 11, N/C Fe 1.81x1014 N (3 SF) 113,000 N/C (3 SF) x10 100,000 N/C (1 SF) 14 N (1 SF) xmple Clculte the spee of proton tht is ccelerte from rest throuh potentil ifference of 10 V q p m p V 10V v? 1.6x x10 C 7 v k 1 W DK mv q q q q m 19 (1.6 x10 )(10) x x10 5 m/s lectric Potentil of Point Chre p to this point we hve focuse our ttention solely to tht of set of prllel pltes. But those re not the ONLY thin tht hs n electric fiel. Rememer, point chres hve n electric fiel tht surrouns them. So imine plcin TST CHARG out wy from the point chre. Will it experience chne in electric potentil enery? YS! Thus is lso must experience chne in electric potentil s well. 4
5 9/3/014 lectric Potentil Let s use our plte nloy. Suppose we h set of prllel pltes symolic of ein ove the roun which hs potentil ifference of 50V n CONSTANT lectric Fiel , V= 5 V 0.5, V= 1.5 V =? From 1 to 5 V =? From to 3 0 V =? From 3 to V =? From 1 to V Notice tht the LCTRIC POTNTIAL (Volte) DOS NOT chne from to 3. They re symoliclly t the sme heiht n thus t the sme volte. The line they re on is clle n QIPOTNTIAL LIN. Wht o you notice out the orienttion etween the electric fiel lines n the equipotentil lines? quipotentil Lines So let s sy you h positive chre. The electric fiel lines move AWAY from the chre. The equipotentil lines re perpeniculr to the electric fiel lines n thus mke concentric circles roun the chre. As you move AWAY from positive chre the potentil ecreses. So V1>V>V3. r V(r) =? Now tht we hve the irection or visul spect of the equipotentil line unerstoo the question is how cn we etermine the potentil t certin istnce wy from the chre? lectric Potentil of Point Chre W F x ; x r q q F r Qq ; F k q r Qqr k k qr Q r Why the sum sin? Volte, unlike lectric Fiel, is NOT vector! So if you hve MOR thn one chre you on t nee to use vectors. Simply up ll the voltes tht ech chre contriutes since volte is SCALAR. WARNING! You must use the sin of the chre in this cse. 5
6 9/3/014 Potentil of point chre Thus theistnce from corner to the center will equl : r V V V center center center Suppose we h 4 chres ech t the corners of squre with sies equl to. If I wnte to fin the potentil t the CNTR I woul SM up ll of the iniviul potentils. Q k r q q 3q 5q k( ) r r r r 5q 10q 5q k( ) k( ) k r lectric fiel t the center? ( Not so esy) If they h ske us to fin the electric fiel, we first woul hve to fiure out the visul irection, use vectors to rek iniviul electric fiels into components n use the Pythoren Theorem to fin the resultnt n inverse tnent to fin the nle resultnt So, ye.lectric Potentils re NIC to el with! xmple An electric ipole consists of two chres q 1 = +1nC n q = 1nC, plce 10 cm prt s shown in the fiure. Compute the potentil t points,, n c. V k q1 q ( ) r r 9 1x10 V 8.99x10 ( 0.06 V 899 V 9 1x10 )
7 9/3/014 xmple cont V k q1 q ( ) r r 9 1x10 V 8.99x10 ( 0.04 V V 9 1x10 ) V c 0 V Since irection isn t importnt, the electric potentil t c is zero. The electric fiel however is NOT. The electric fiel woul point to the riht. 7
I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
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