Electrostatic Forces and Fields

Transcription

1 Electotatic Foce and Field oulomb Law Electic chage i an intinic popety of matte, in exactly the ame fahion a ma. At the atomic level electic chage come in thee type and i caied by the thee elementay paticle. Thee chage ae the poton (a poitive electic chage), the electon (a negative electic chage), and the neuton (a zeo electic chage.) Thee elementay paticle ae the building block fo all matte and typical atomic tuctue a evealed by expeiment ha a heavy dene nucleu (compoed of poton and neuton) uounded by electon in pecific pobability ditibution, o electon cloud called allowed obital. Atom in geneal ae electically neutal meaning that the numbe of poton i equal to the numbe of electon, but ince the electon ae looely bound the nucleu, they may be eaily added to o emoved fom the atom. Ionization of an atom i the poce by which electon ae added o emoved with minimal input of enegy. If electon ae emoved fom an electically neutal atom, a negative ion i fomed. onveely, if electon ae added to an electically neutal atom, a poitive ion i fomed. Expeiment with electic chage demontate a numbe of phenomena. The fit of which i that the magnitude of the chage on the poton and the electon ae the ame and thi value, called the mallet amount of fee chage found in natue, i found to be that of the chage on the electon. Thu one elementay chage ha a magnitude of p 19 = e = e = 1.6. Thee ae eveal implication that can be made hee, and can be een to hold tue expeimentally. The fit i that electic chage i quantized, o

2 occu in only dicete quantitie, namely total = n e, whee n i the numbe of dicete elementay chage, e, that make up the total chage total. The econd implication that can be made i thee could be a collection of moe elementay chage that exit only in bound tate and thee could make up an elementay chage. Fo example, it i found that quak ae fundamental paticle that ae the contituent paticle out of which that make up poton and neuton ae made. uak exhibit factional electic chage in unit of 1 ± e o ± e and neve occu a a ingle fee unit of chage. uak only occu a pat 3 3 of a bound ytem. In addition, it i expeimentally found that if two chage 1 and, cay the ame algebaic ign, the chage will be epelled away fom each othe, while if they cay the oppoite algebaic ign they will be attacted. Thi implie that thee mut be a foce that exit between electic chage. Thi fom of thi foce law wa expeimentally detemined by oulomb in the 1870 and bea hi name. oulomb expeiment involved placing amount of chage, 1 and on two iolated inulating phee epaated by a known ditance. A chage 1 wa placed on a dumbbell that wa fee to otate about a wie though the cente of the dumbbell. The econd inulating phee with chage wa bought cloe to 1 and would caue 1 to otate about the wie. The toque that wa ceated i elated to the magnitude of the electotatic foce that exit between 1 and. oulomb found two impotant eult fom hi expeiment. The fit, given a fixed amount of chage 1 and on each of the phee, the magnitude of the electotatic foce vaied a the invee of the quae of the ditance between the cente of the two chage. Second, fo a fixed ditance between the chage, the magnitude of the electotatic foce vaied a the poduct of the two chage on the

3 1 phee. oulomb eult can be ummaized a follow: Felectota tic, whee 1, i the cente to cente ditance between 1 and. To make thi eult an equality athe than a popotionality, a contant i multiplied on the ight hand ide, and thi contant of 9 Nm popotionality i called oulomb contant and ha a value of8.99. Thi value can be expeimentally veified by eceating oulomb expeiment a i ometime done in undegaduate phyic laboatoy clae, keeping eithe the chage fixed (and vaying the ditance between the chage) o keeping the ditance fixed (and vaying the amount of chage placed on each phee.) In eithe cae, gaphical analyi will then yield a value fo oulomb contant. Theefoe, oulomb Law fo tatic ditibution of 1, chage i geneally witten a F electotatic 1 = k 1, ˆ, whee, ˆ i a unit vecto that point along the line joining the two chage. In othe wod, oulomb Law give the magnitude of the foce that exit between the two chage and alway point along the line joining the two chage. If thee i moe than one chage peent, oulomb Law till applie and the electotatic foce on one chage due to all othe chage i the vecto um of all of the foce on that chage due to the othe chage peent. A one final comment, oulomb Law i applicable to point chage (object that ae much malle than the ditance between them) that ae at et (tatic.) Example #1 How big i a oulomb of hage? Suppoe that a two inulating ball ae epaated by 1.0 m and that a chage of 1.0 i placed on each. What i the magnitude of the electotatic foce felt by one of the phee?

4 Solution Applying oulomb Law we find 1 9 Nm F = k = N =. 1, ( 1.0m) Thi equate to a weight of about.x 9 pound, given that a weight of 1 N ~ ¼ pound. Hee we can daw the concluion that one oulomb of chage i a huge amount and in geneal we will only ee faction of oulomb of chage in pactice. In othe wod we will in geneal ee mico-oulomb (1µ = 1x -6 ) o nano-oulomb (1n = 1x -9 ) of chage. Example # How many elementay chage ae in 1 µ of chage? Solution We take 1 µ of chage and divide thi by one elementay chage. 1e 1 = = 6.5 e # e 1µ oulomb Law fo the electotatic foce i alo a Newton Thid Law foce. To ee thi, we conide the aangement of chage hown below and explicitly wite out the foce that act on each chage. 1 > 0 > 0 F,1 F 1, 1 > 0 F,1 F 1, < 0

5 Fom thi diagam we ee that fo any two chage, the foce exeted on the chage ae equal in magnitude and oppoitely diected and thu we have F 1, = F, 1, which i Newton Thid Law. Ealy on in the dicuion of electic chage we aid that thee ae thee type of chage. Let uppoe that we have a poitive electic chage (a poton) that i epaated fom a negative electic chage (an electon) a would be typically found in a hydogen atom, and let u calculate the electotatic foce felt by eithe the poton o the electon. Example #3 What i the magnitude of the electotatic foce between a poton and an electon in a hydogen atom? Solution To calculate the magnitude of the electotatic foce that exit between a poton and an electon in a hydogen atom, we need to know the epaation of the two chage. Typically fo a hydogen atom in it gound tate, we have the epaation between the poton and the electon, known a one Boh adiu, 0.53x - m. Applying oulomb Law we find fo the magnitude of the electotatic foce F electotatic 1 = k 1, e e = k = p, e Nm 19 ( 1.6 ) ( 0.53 m) = 8. 8 N. The diection of the electotatic foce that the electon feel due to the poton i diected towad the poton and ha a magnitude of 8.x -8 N, while the electotatic foce that the poton feel due to the electon i diected towad the electon in accodance with Newton Thid Law, and ha a magnitude alo of 8.x -8 N.

6 In geneal when we do poblem involving foce we uually have to woy about the gavitational foce exeted on the object. So let ak when dealing with elementay paticle do I have to woy about gavitational foce? In othe wod, do I have to woy about the weight of thee paticle? To anwe thi quetion, we will calculate the magnitude of the gavitational foce between the poton and the electon in a gound-tate hydogen atom and compae thi to the electotatic foce calculated above. Example #4 The gavitational foce between the poton and electon in a hydogen atom Solution - To calculate the magnitude of the gavitational foce we need to ue M1M Newton Univeal Law of Gavity F gavity = G 1,. Looking up the mae of the poton and the electon we find M p = 1.67x -7 kg and M e = 9.11x -31 kg epectively. Uing the epaation between the poton and the electon in a hydogen atom, we find the magnitude of the gavitational foce to be F gavity Nm 1.67 kg 9.11 kg 47 =.67 = 3.6 N. 6 kg ( 0.53 m) ompaing thi to the electotatic foce given above we ee F that F electotatic gavity =.3 39, which implie that the electotatic foce i much lage than the gavitational foce, by a facto of 39. Thu we don t need to woy about the foce of gavity when woking with elementay paticle. Thi i not tue when we have othe moe maive object, like coffee cup, ball, and aiplane!

7 Now, what would happen if we wee to omehow eleae the poton and the electon and allow them to move unde the influence of the electotatic foce, what would happen? Example #5 olliding a poton and electon Suppoe that a poton and an electon ae epaated by a ditance d (maybe d could be the typical epaation in a hydogen atom.) Suppoe futhe that the poton i at the oigin and that the electon i to the ight of the poton by thi ditance d. If the electon and poton ae eleaed fom et at the ame time, qualitatively whee will the poton and the electon collide? Solution The poton and the electon will expeience the ame magnitude of the electotatic foce. Howeve, ince the poton i moe maive than the electon, by about a facto of 1881, and thu the acceleation of the poton and the electon will be diffeent. So, let u fit calculate the initial acceleation of the poton and the electon. Fom Newton Second Law, the acceleation of the poton i F 8. N a = 4.6 kg 8 p = = 7 mp m, while the acceleation of the electon i F 8. N a = 9.0 kg 8 e = = 7 me 9.11 m. Since the acceleation of the electon i geate than that of the poton, in the ame inteval of time, the electon will have a geate change in velocity and cove a lage ditance than the poton. Theefoe, the poton and the electon will collide cloe to the poton at a ditance omewhee between 0 and d/. Be aeful Thi acceleation i not a contant of the motion ince the foce i not contant, but change with ditance. You would not want to ue the claical equation of motion (valid fo contant acceleation) to calculate the poition of the colliion.

8 Equation of motion fo thi non-contant acceleation could be developed but ae beyond the cope of thi text and unfotunately only a qualitative olution can be given at thi point. So fa we have een oulomb Law and ome example of how to calculate magnitude of the electotatic foce that exit between two object with chage. Now we will tun ou attention to ome moe ophiticated poblem and exploe the vecto natue of oulomb Law in which we apply it to ituation involving moe than two chage. To do thi we will need a tategy. My tategy i a follow and thi methodology will be ued to olve all of the poblem that involve vecto. Fit you need to pick a convenient coodinate ytem. It doe not matte what that coodinate ytem i, but the choice hould be well uited to the phyical ituation and you need to be conitent when aigning algebaic ign to the vecto quantitie baed upon thi coodinate ytem. Next, I will daw all of the vecto that epeent the phyical quantity of inteet on the object of inteet. Typically thi mean that I will pick a chage and daw all of the foce, ay, that act on that chage. I will then beak up thoe phyical quantitie epeented by the vecto into thei component (baed on the choice of coodinate ytem) and um the vecto algebaically to calculate the net component aociated with a paticula phyical quantity in a paticula diection. I will then epot the eult a a vecto (uing unit vecto notation) o a a magnitude and a diection (meaued with epect to ome convenient tating point.) Alight, that a whole bunch of wod. How do you actually apply thi method to olve poblem? To anwe that, we will look at two pecific example. The fit i a

9 one-dimenional poblem involving thee chage in a line and the econd i a twodimenional poblem involving chage located on the vetice of a ight-tiangle. Example #6 Thee chage in a line What i the net electotatic foce on the leftmot chage hown below due to the othe two chage, whee 1 = -8 µ, = 3 µ, and 3 = -4 µ? Solution - Given the diagam below we chooe the oigin of the coodinate ytem to be at the leftmot chage and elect to the ight a the poitive x-diection. 0.3m 0.m x -8µ 3µ -4µ Dawing the foce on the leftmot chage we can apply oulomb Law and wite F 1,3 F 1, -8µ F F net,1 net,1 = F 1,3 + F = 9 The ymbol 1, 9 Nm F A, B 1 = k 1, k 1, iˆ ( 0.5m) ( 0.3m) = + 1.Niˆ ued hee mean the electotatic foce on object A due to object B. Thu the net electotatic foce on the leftmot chage i 1. N diected along the poitive x-axi.

10 Example #7 Thee chage at the vetice of a ight tiangle What i the net electotatic foce on the 65 µ chage due to the othe two chage hown, whee 1 = 65 µ, = 50 µ, and 3 = -86 µ? y 65µ 0.30m 0.6m 50µ 0.5m 30 o -86µ x F 1, 65µ F 1,3 The foce on the 65 µ chage ae a hown above and we need to find the net x- and net y-foce. Thu fo the net x- and y-foce we have F F net, x net, y 31 : k coθ = 9 3,1 1 : + k,1 9 Nm k inθ = 9 3,1 6 9 Nm 50 ( 0.6m) co30 = N ( 0.3m) ( 0.6m) in 30 = 60N Thi give uing the Pythagoean Theoem the net electotatic foce a 60N FNet = 65 N 1 o ( N ) + ( 60N φ = tan = φ =

11 Now let apply what we ve leaned o fa to omewhat moe complicated example. Example #8 How much chage doe it take? The leave of an electocope ae contucted out of two identical ball whoe ma ae 3.x - kg and thee ball hang in equilibium fom two ting of length L = 15 cm. When in equilibium, the ball make a 5 o angle with epect to the vetical. How much chage i needed on each phee to poduce thi ituation? Solution We chooe to examine the foce on the left mot chage and aume a tandad ateian coodinate ytem. The phyical epeentation of the poblem tatement i hown below left. The foce that act on the ma ae the tenion foce in the ting, the weight of the ma, and the electotatic foce and ae hown below ight. 5 o F E m, F T 5 o m, m, F W Applying Newton nd Law in the hoizontal and vetical diection we find F : F x F : F y T T in 5 F E co5 F W = max = 0 FE = k = FT in 5 mg = may = 0 FT = co5

12 So, we have two equation and two unknown, namely F T and (fom F E.) Subtituting F T fom the ummation of the y-foce into the expeion in the x-foce, we can olve fo. Thi poduce = mg in 5 k co5 = ( Lin 5) mg in 5 k co5 = fo the chage needed on each phee. In ode to pefom the above calculation, we have alo ued the fact that the chage ae epaated by a ditance, one-half of which can be calculated fom the angle given and the length of the ting. Thu we have = Lin 5 = Lin 5. In addition we could alo calculate the magnitude of the m mg 3. kg 9.8 tenion foce, FT = = = 0. 3N co5 co50 38 Example #9 The adioactive decay of U. In the adioactive decay of uanium U, the cente of the emeging alpha 4 paticle He i, at a cetain intant 9x -15 m fom the cente of the thoium daughte nucleu Th. a. What i the magnitude of the electotatic foce on the alpha paticle? (Hint: In A pectocopic notation, Z X A epeent the atomic ma in atomic ma unit (amu), Z epeent the atomic numbe, and X i the element.) Solution - The magnitude of the electotatic foce i given by oulomb Law and i due to the electotatic epulion of the poton that make up the alpha paticle and the thoium nucleu: Th 9 α Nm ( e ) ( 90e ) Fα, Th = k = 9 51N = 15 α, Th 9 m ( ) b. What i the acceleation of the alpha paticle? (Hint: 1 amu = 1.67x -7 kg.) Solution - The acceleation of the alpha paticle i found fom Newton nd Law: 51N 51N 8 m F α = mα aα = 51N aα = = = m kg α

13 The Electic Field Mot Newton Law foce ae called contact foce becaue fo an object to acceleate becaue of an applied foce, the foce ha to be applied to the object though diect phyical contact. Howeve, ome foce, like Gavity and the Electotatic foce extend ove ditance even when the object do not touch. Thee ae called action at a ditance foce. Foce like thee ae exeted on mae (due to the Foce of Gavity) o chage (due to the Electotatic Foce) though a field. To invetigate thee field we ue a tet object. Fo gavitational field, we ue a tet ma m o while fo electic field we will ue a tet chage q 0. Hee we will aume that the tet chage i poitive and vey mall in ma compaed to the chage whoe electic field we want to invetigate. The choice of q 0 being poitive i a convenience and choen to mio the eult you get fom tudying the motion of mae in a gavitational field. The fact that the tet chage i much le maive i o that the acceleation of the tet chage i much lage than the chage whoe field i being invetigated. Since both chage exet equal and oppoite foce on each othe, I would like only the tet chage to move and have the field due to the othe chage be tatic. Since the electotatic foce will caue q 0 to move, if i a poitive chage, then q 0 will feel a epulive foce fom and move away. If, howeve, i a negative chage, then q0 will feel an attactive foce and move towad the chage. The electic field line will then point adially outwad fom a poitive chage and point adially inwad towad a negative chage a hown below.

14 Example # - The Electic field line due to a poitive and negative electic chage invetigated uing a mall poitive tet chage q 0. > 0 < 0 Electic field line ae pictoial epeentation of the electic field that exit in pace due to the peence of a chage. In othe wod they epeent the line of foce felt by q 0 due to the inteaction of q 0 with the electic field due to. Thu the electotatic foce on q 0 due to i exeted though q 0 inteaction with the electic field of. The electic field uound the chage and exit whethe o not a tet chage i peent to invetigate the field. Thi i exactly analogou to a gavitational field. A gavitational field exet a foce on a tet ma m o though m o inteaction with the gavitational field of a ma M and the gavitational field of M exit even if thee i no tet ma peent to pobe thi field. In addition, the electic field line indicate the diection of the electic field at any point in pace due to ome ditibution of chage. The magnitude of the electic field, a vecto quantity, ha to be detemined by the upepoition of all the electic field due to all chage peent. WE define the electic field of a point chage a F the electotatic foce pe unit tet chage, o E =, o that the electotatic foce may q o

15 v be witten a F = q E. Thu fo a point chage, we have fo the magnitude of the o electic field E = k and the diection i tangent to the field line at that point in pace. Electic field line can be een to tat on poitive chage, extending adially outwad, and end on negative chage, pointing adially inwad. The numbe of electic field line (pe unit aea of pace) i popotional to the magnitude of the electic chage peent. In othe wod, the flux (o numbe) of electic field line though an aea pependicula to the electic field vecto i popotional to the total chage encloed by a uface of aea A. Thi i called Gau Law and i ued to calculate the electic field fo an abitay ditibution of chage. Gau Law can be witten a Suface E da = ε encloed o and i ueful when tudying the popagation of electomagnetic adiation, o light. We will apply Gau Law to a couple of ituation below, only a illutative example, even though we will only eve ue electic field of a point chage E = k a given above. Example #11 What ae the electic field due to a point chage, a unifom plate of chage (pe unit aea), and a unifom line of chage (pe unit length) uing Gau Law? Solution - Fo a point chage, we chooe to have the vecto pointing in the diection of the electic field due to a poitive chage a hown above. Fit, we find that at a fixed ditance away,, the flux of field line i contant, indicating that the electic field i a contant. Evaluating Gau Law we have

16 Suface E da = E da = 4π phee of adiu E = ε encloed o E point chage = 4πε o = k, which vaie a the invee of the ditance quaed fom the point chage. - Fo a plate of chage pe unit aea we define σ = / A, and choe the aea vecto to be in the diection of the electic field a hown ight. Evaluating Gau Law we have Suface E da = E plate da = σa = = ε Aε encloed EA = Eplate o and independent of how fa you ae away fom the plate. o σ, which i contant ε - Fo a line of chage pe unit length we define λ = / L, and choe the aea vecto to be in the diection of the electic field which point adially out fom the line of chage a hown below. Evaluating Gau Law we o have Suface E da = E da = line of chage encloed λl λ πle = E = = k, which vaie a ε ε πl line o the invee of the ditance fom the line of chage. o Retuning to point chage, we will ty to evaluate the electic field at a point P in pace given ome diffeent ditibution of chage. To do thi we will ue the method outlined in the ection on oulomb Law fo olving vecto poblem.

17 Example #1 What i the electic field at point P (located at a ditance of 0.4m on the y-axi) due to the two chage ( 1 = 7 µ and = -5 µ) located on the x-axi and epaated by 0.3 m? Solution Given the diagam below, we daw the electic field due to the two chage at point P. y E 1 P 0.4m E 1 0.3m x Beaking thee electic field vecto up into thei x- and y-component we find E E net, x net, y = k, p = + k 1 1, p coθ = 9 k, p 9 Nm 5 inθ = 9 ( 0.5m) 9 Nm 6 0.3m = m ( 0.4m) ( 0.5m) 5 N 0.4m = m Uing the Pythagoean Theoem we can expe the net electic field at point P a 5 N. E E net net = 5 N 5 N ( 1.08 ) + (.50 ) =.7 5 φ = 79.4 φ = tan N N.

18 Example #13 What i the electic field at a point P (along the poitive x-axi) due to the electic dipole ituated on the y-axi. Solution Fit an electic dipole i contucted by two equal and oppoite magnitude chage epaated by a ditance d between thei cente. Hee the ditance between thei cente i d = a, whee a i the ditance the chage i located above and below the x- axi. Second, we will need the ditance between the chage and whee we want to evaluate the electic field. Thi ditance,, i the ame fo both chage and i given a = a + x. Thid, given the coodinate ytem in the diagam, we daw the electic field at point P due to the two chage located on the y-axi. Then we beak up the field into thei x- and y-component and calculate the net electic field in the x- and y- diection. y a -a x θ E P E + x In component fom we find fo the net electic field in the x- and y-diection E E net, x net, y = + k +, p = k, p coθ k, p inθ k +, p coθ = 0 N inθ = k, p inθ = k a 3 ( a + x ). Thi give the net electic field in the negative y-diection with magnitude k a 3 ( a + x ). In fact thi ame analyi could be done anywhee along the y-axi and the eult will be

19 the ame magnitude and will point along the negative y-axi. If we take the limit that the a a point P i at a ditance x >> a, the dipole field educe to k k and the field 3 3 x fall off a the invee of the ditance cubed! In the above two example we have been given a ditibution of point chage and wee aked to calculate the electic field at a point P in pace. A a lat couple of example in thi ection, let take a chage and ee what happen a the chage expeience an electic field. Example #14 The electic field needed to balance a poton againt gavity What i the magnitude and diection of the electic field needed to balance the weight of a poton in the gavitational field of the eath? Solution The weight of the poton point vetically downwad towad the eath and ha a magnitude of mg. To balance the weight of the poton, the electic foce mut point vetically up, and ince the poton caie a poitive electic chage, the electic field mut alo point vetically up. Thu we have pointing vetically up, an electic field of magnitude F E F W = ma y = 0 F E = qe = mg E = mg q kg 9.8 = m = N. Example #15 Pojectile motion with an electon? Suppoe that an electon ente a egion of unifom electic field (E = 00 N/) with an initial velocity of 3x 6 m/. What ae the acceleation of the electon, the time it take

20 the electon to tavel though the field and the vetical diplacement of the electon while it i in the field if the length of the egion L = 0.1m? Solution Fom the diagam of the et-up below, we ee that a foce will acceleate the electon vetically down ince the electon ha a negative chage. Thi acceleation will be contant, ince we ae told that the electic field i unifom, and thi poduce a contant foce. Thu the acceleation of the electon i vetically down with v E L 19 N F qe m magnitude a = = = = Thi foce only affect the m m 9.11 kg vetical motion of the electon and the hoizontal component of the velocity i theefoe a contant of the motion. The vetical component of the velocity will change with time. The time it take then, fo the electon to tavee the length of the plate i given at = L v x.1m = = m In thi time the electon i acceleated downwad towad the poitive plate. Defining yi to be zeo whee the electon ente, the vetical diplacement i y a t ( ) m 8 = = 1.95 m below whee the electon enteed. f = o 1.95cm y 3 omment: Hee we have ued the equation of motion developed in the mechanic potion of the coue valid fo motion with contant acceleation. If we ignoe the plate a the caue of the foce on the electon and futhe petend that the electon wee not

21 chaged, but imply a ma m moving unde the influence of gavity, we have a cae of two dimenional motion, o moe pecifically pojectile motion, hee with an electon in an electic field. Example #16 What wa the wok done on the electon in example #15 and doe thi equal the change in kinetic enegy of the electon? Solution We will tackle thi poblem in two pat; fit we will calculate the wok done and then we will calculate the final velocity fom the equation of motion and compute the change in kinetic enegy fo the electon in thi cae. So, the wok done by the electotatic foce in acceleating the electon i given a W E = F y = F y coθ = qe y = N 1.95 m = 6.4 Fom Example 15, we know that the hoizontal component of the electon velocity i 6 m contant, o v = 3.0. The final vetical component of the electon velocity fx 19 J. i v fy 8 = a t = = 1.17 y 13 m 6 m. Uing the Pythagoean Theoem, the final magnitude of the electon velocity 6 m i ( ) ( ) m 6 m 6 v f v fx + v fy = = 3. =. To ee if the wok done equal the change in kinetic enegy we need to compute the change in kinetic enegy. The change in the electon kinetic enegy i m 6 m ( v ) = 9.11 kg ( 3. ) ( 3.0 ) 19 ( ) = 6. J KE = 1 m v f i which eem to agee petty well. Thu, the wok done i equal to the change in the electon kinetic enegy a it hould by the wok-kinetic enegy theoem.,

22 Thi lat example bing u to ou next topic, the wok done in aembling o moving chage aound in the electic field of othe chage. We ae tating to bidge the pace between having puely tatic chage and finally allowing thee chage to move, thu etablihing electic cuent and the exitence of magnetic field and foce. In the next chapte we will examine the concept of the electic potential and of electic potential enegy. We will tat by looking at gavity and ecalling what we know about how conevative foce give ie to coneved quantitie like enegy. Then we will, by analogy with gavity, dicu ome fundamental concept in ou tudy of electicity.