CONDUCTORS AND INSULATORS

Transcription

1 CONDUCTORS AND INSULATORS We defne a conductor as a materal n whch charges are free to move over macroscopc dstances.e., they can leave ther nucle and move around the materal. An nsulator s anythng else. In an nsulator the charge dstrbuton n an atom may change, but the charges do not leave ther nucle. When we consder electrostatcs, the case where charges are not movng, ths defnton has several mmedate consequences. 1. There can be no electrc feld nsde a conductor. If there were, t would exert a force on the charges causng them to move. They would adjust untl there was no feld. 2. There can be no net charge at any pont nsde a conductor. Gauss s Law would gve E da 04kQQ0 3. The electrc feld at the surface of the conductor must be perpendcular ( ) to the surface. If there were a component parallel to the surface t would cause charges to move along the surface untl there was none. There are other consequences we wll nvestgate later. Symmetres Gauss s Law s always true, but n sum form t s generally only useful f we can get E outsde the surface sum. Ths generally requres a specal symmetry. We wll examne three such symmetres. Sphercal Symmetry Consder the followng arrangement.

2 It conssts of two hollow concentrc conductng shells as shown. A charge Q 0 s placed at the center, Q 1 on the nner shell, and Q 2 on the outer shell. The rest of the space s a vacuum. Fnd the electrc feld and charge densty everywhere. r R 1 Draw an magnary sphere of radus r as shown. Because of the sphercal symmetry we know that the magntude of E s the same everywhere on the sphere of radus r, and also that t s radal (ponts n ˆr drecton). Thus. kq E da E da cos(0) E da E 4 r 4 kq E rˆ r R1r R2 Agan, draw an magnary sphere n ths regon Snce the sphere s nsde the conductor we know E = 0. But then Gauss s Law tells us the total charge nsde the sphere s 0. Ths conssts of Q 0 and Q where Q s the charge on the nner 1 n 1 n

3 surface of the conductor. Snce there can be no charge n the conductor, and the total charge on the conductor s Q 1, we must have Q Q Q out Q 1 0 n Q E 0 R2 r R3 Ths s the same as the 1 st regon wth Q 0 replaced by Q 0 + Q 1 Ths s the same as regon 2 Ths s the same as regons 1 and 3 k Q 0 Q E 1 r ˆ 2 r out R3 r R4 Q Q Q Q Q Q Q n E 0 r R 4 k Q 0 Q 1 Q E 2 r ˆ 2 r We see that the felds are ether 0 or the same as f all charges were at the center. Cylndrcal Symmetry Next we consder a conductng cylnder of radus R and length L carryng a charge Q 0.

4 Because of the cylndrcal symmetry we know that the feld magntude must be ndependent of θ (the angle about the axs of the cylnder). Further we know that E must be to the surface of the cylnder. We do not know that t has to stay that way. However, at center t must be by symmetry. To see what happens at other ponts we note that f we go far away from the cylnder compared to R or L, t would appear as a pont charge. Hence the feld wll appear to be radal at large dstances. Snce t must start to the cylnder t must look lke

5 Thus f we stay close to the cylnder (r << L) the lnes wll be to the cylnder (drected along the radus of the cylnder). If we use cylndrcal coordnates wth ẑ along the axs ths becomes E E,zˆ The remanng queston s the z dependence. We expect the charge densty to be greater near the ends than at the center because lke charges repel (at the rght hand end there s no charge pushng back, to the left, hence charge wll tend to accumulate there). However, f we are near the mddle we can t see ether end and hence there should be no dfference n charge densty f we look toward ether end. Ths means t must be unform. Thus f we stay away from the ends we expect unform charge densty Q ~ R 2 L We wll thus approxmate the stuaton as one n whch E does not depend on z. Then draw an magnary cylnder of radus r as shown

6 Then E da E da cose 2r 4kQ L n 2kQ E 0 ˆ Lr Note that the feld falls off as 1/r rather than 1/r 2. Ths s because the lnes spread out n one dmenson rather than two. Ths same phenomenon has nterestng mplcatons n modern strng theores of more than 3 dmensons. Next consder the analogous problem to the concentrc conductng spheres. R 1 2kQ E 1 ˆ Lr R Q R 1 2 out 1 0 n Q Q Q Q R E 0 R 2 3

7 E 2k Q0 Q1 ˆ R n Lr R 3 4 Q Q Q out Q Q Q Q E E 0 R 4 2k Q0 Q1Q2 ˆ Lr Planar Symmetry Fnally, we consder the case of a plane carryng a charge Q 0. We do ths just as we dd for the cylnders. The charge wll tend to cluster near the edges, but f we stay away from them t wll be unform. The feld wll start to the surface and then bend toward the edges when the dstance from the sheet becomes comparable to L. Hence, away from the edges and close to the sheet we draw a rectangular box

8 By symmetry the feld s the same above and below the sheet. Thus a E da E da cose 2a 4kQ A 0 Q E 2 k nˆ 2 k nˆ A where σ s the surface charge densty and ˆn s the normal to the surface. Note that ths tme the feld does not fall off wth dstance obvously because the lnes don t spread!!