PES 110 Sprng 014, Spender Lecture 6/Page 1 Lecture today: Chapter 1) Electrc feld due to charge dstrbutons -> charged rod -> charged rng We ntroduced the electrc feld, E. I defned t as an nvsble aura surroundng every charge. We saw that E s a vector feld (we assgned a vector to every pont n space). The vector E tell us n whch drecton and wth what magntude a test charge q feels an electrostatc force The electrc feld tells us that a postve test charge wll always move away from a postve source charge at the orgn. We also know how to calculate the electrc feld when there are more than one pont charges. We smply add them by the prncple of superposton: E r E1 r E r... Now wll consder the case of contnuous rather than dscrete charges. So far we have only consdered one or two or three pont charges. In many stuatons charge s spread over an area or throughout a volume of space so we nstead talk about charge densty. Charge Dstrbutons The electrc feld created by charge dstrbutons are completely dfferent from those of pont charges. Charge Dstrbutons s created by nfntely many charges spread over a surface or throughout the volume of an object. There are three dfferent types of charge dstrbutons: Lne, Area, and Volume dstrbutons. Volume Charge Densty Suppose we wsh to fnd the electrc feld at some pont P. Let s consder a small volume element V whch contans an amount of charge q charges wthn the volume element than compared to r, the dstance between. The dstances between V are much smaller V and P.
PES 110 Sprng 014, Spender Lecture 6/Page In the lmt where densty r as V becomes nfntesmally small, we may defne a volume charge r q lm V V 0 dq dv unts [C/m 3 ] The total amount of charge wthn the entre volume V s Q q r dv V NOTE: The concept of charge densty here s analogous to mass densty m. When a large number of atoms are tghtly packed wthn a volume, we can also take the contnuum lmt and the mass of an object s gven by M m r dv V Surface Charge Densty In a smlar manner, the charge can be dstrbuted over a surface S of area A wth a surface charge densty σ (lowercase Greek letter sgma): dq r unts [C/m ] da The total charge on the entre surface s: Q r da S Lne Charge Densty If the charge s dstrbuted over a lne of length, then the lnear charge densty λ (lowercase Greek letter lambda) s dq r unts [C/m] dl total charge on the entre surface s: Q r dl lne NOTE: If charges are unformly dstrbuted throughout the regon, the denstes (ρ, σ, λ) then become unform. r
PES 110 Sprng 014, Spender Lecture 6/Page 3 Example 1: A charged non-conductng sphere, wth a radus of.0 cm, has a non-unform volume charge densty of ρ(r) = cr, where c = -1.5 μc/m 4. What s the total charge on the sphere? Electrc Felds due to Contnuous Charge Dstrbutons We know that for a pont postve source charge q located at the orgn, the electrc feld surroundng the source charge s 1 q E r rˆ 4 0 r Smlarly, the electrc feld at a pont P due to each charge element dq s gven by Coulomb s law: 1 dq de r rˆ 4 0 r where r s the dstance from dq to P and ˆr s the correspondng unt vector. (See Fgure above). Usng the superposton prncple, the total electrc feld E s the vector sum (ntegral) of all these nfntesmal contrbutons: E r 1 4 0 V dq rˆ r Ths s an example of a vector ntegral whch conssts of three separate ntegratons, one for each component of the electrc feld. Then we use the charge denstes to replace dq: 3D: dq r dv D: dq r da dq r dl 1D:
PES 110 Sprng 014, Spender Lecture 6/Page 4
PES 110 Sprng 014, Spender Lecture 6/Page 5 MORE ON Charged ROD: We just calculated that: ll 1 E k d L / 4 d In the lmt where d >> L, the above expresson reduces to the pont-charge lmt: ll 1 ll 1 ll Q E k k k k d L / 4 d d L / 4d 1 d d Ths makes sense snce a fnte lne looks lke a pont at a very far dstance!
PES 110 Sprng 014, Spender Lecture 6/Page 6 WORKSHEET: Integratng to fnd the electrc feld due to a charge dstrbuton
PES 110 Sprng 014, Spender Lecture 6/Page 7 MORE on Charged RING: zl( pr) zq zq E k k k 3 z R z R z 1 R / z 3/ 3/ 3/ Let us check ths equaton for a pont on the central axs that s so far away that z >> R. For such a pont, the expresson becomes: zq E k k 3 z Q 1 / 3/ z R z Ths s a reasonable result because from a large dstance, the rng looks lke a pont charge. Let us next check the result for a pont at the center of the rng that s, for z = 0. At that pont, the expresson tells us that E = 0. Ths s a reasonable result because f we were to place a test charge at the center of the rng, there would be no net electrostatc force actng on t; the force due to any element of the rng would be canceled by the force due to the element on the opposte sde of the rng. If the force at the center of the rng were zero, the electrc feld there would also have to be zero.