Lecture 15 - Current Puzzle... Suppoe an infinite grounded conducting plane lie at z = 0. charge q i located at a height h above the conducting plane. Show in three different way that the potential below the conducting plane i zero everywhere. 1. We can conider the pace z < 0 to be inide of a giant hemipherical conductor which i bounded by z = 0 from above and by the infinite hemiphere everywhere below and around it. The potential on thi entire conductor i 0. One obviou olution i that the potential everywhere in thi region i zero, and the Uniquene Theorem guarantee that thi i the only olution.. In your book (more preciely, Figure 3.1), the cae of a finite (ungrounded) circular dik conductor wa dicued. In uch a cae, a total charge + q ditribute itelf (roughly) evenly on the underide of the conducting heet. In the limit a the radiu R of the dik goe to infinity, we recoup the infinite conducting plane, which would have charge denity zero. lthough thi heet i not grounded, it clearly ha potential zero, and therefore grounding it will not change the charge ditribution. Therefore, there will be zero charge denity, and therefore zero potential, everywhere for z < 0. 3. Imagine that the infinite conducting plane i actually a conducting pherical hell with radiu R and center (0, 0, -R) in Carteian coordinate. The limit R would yield the ame geometry a the conducting plane, and the electric potential inide of the grounded pherical cavity would be zero everywhere (thi i the unique olution to Laplace equation with ϕ = 0 everywhere on the boundary). lternatively, you can imagine a phere with q radiu R and center (0, 0, R) which contain the point charge. In thi cae, there would be a charge ditribution uniformly acro the urface of the phere, which goe to zero a R, and therefore lead to zero potential 4 π R everywhere outide the phere (which correpond to z < 0 in the cae R ). 4. We could ue an image charge to olve thi problem. Recall that image charge are alway located outide of region() where the potential i to be calculated. Therefore, we are free to ue any image charge we want in the region z > 0. One wie choice would be to put a point charge -q right on top of the charge +q, which would cancel all of the charge and yield potential zero everywhere. Thi reult (which i only valid for z < 0), tell u that the potential below the conducting plate i zero everywhere. dvanced Section: Image Charge for Sphere Image Charge for a Grounded Spherical Shell charge Q lie at a ditance outide of a grounded conducting phere of radiu R. Prove that placing an image charge -q = - Q R R at a ditance from the center of the phere (along the ame line a the charge Q) yield a contant potential 0 everywhere on the phere. Ue thi to determine the force felt from the charge Q from the conducting phere.
Lecture 15-03-06-017.nb Let u determine the potential at point P, which lie at an angle θ relative to the line connecting the center of the phere to the image charge. The ditance from the charge Q to the point P equal while the ditance from -q to P equal Therefore, the potential at point P equal k Q r Q + k (-q) r -q = r Q = (R + - R Co[θ]) 1/ (1) r -q = R + R - R3 Co[θ] 1/ () k Q R + - R Co[θ] 1/ - k Q R R + R - R3 Co[θ] 1/ = 0 (3) Since θ wa arbitrary, thi how that the potential i 0 acro the entire phere. What an incredible reult! With thi in hand, we can now eaily calculate the force on the charge Q ince it equal the force that the image charge -q would exert upon Q in the diagram above, which equal x F = k(-q) Q - R = - k Q R Q x - R = - k R Q x -R a deired. There are many exciting limit to check. For example, Exercie 3.46 examine the limit when the real charge i very cloe to the grounded phere, in which cae the etup mut reduce to the image charge problem of a grounded infinite plane. Bump on a Plane n infinite conducting plane ha a hemipherical bump on it with radiu R. point charge Q i located a ditance R above the top of the hemiphere. Uing the reult of the previou problem, find the image charge needed to (4)
Lecture 15-03-06-017.nb 3 top hemiphere. Uing previou problem, image charge make the electric field perpendicular to the plane and the hemiphere at all point. Recalling the reult of the previou diagram, where = R, we can try putting an image charge - Q R = - Q R at a ditance = R above the center of the hemipherical bump. But that will not be enough to enure that the plane i equipotential; in order to do that we need to add two identical but oppoite charge below the plane. But wait, don t the two extra charge that we jut inerted me up the potential on the hemiphere...? No, they don t! Becaue by the reult of the previou problem, the bottom two charge (a well a the top two charge) caue 0 potential all over the phere. Furthermore, the inner two charge and the outer two charge caue 0 potential all over the plane. So in face, we have created an equipotential urface coniting of the union of the plane and the phere (i.e. the olid and dahed urface in the diagram above), even though we don t care about the bottom of the hemiphere or the equatorial dik inide the phere (i.e. the dahed line). Thi olution can be eaily generalized o that the charge Q i any ditance above the hemipherical bump (but it won t work for a charge Q inide of the hemipherical bump, ince you could not etup image charge only in the outide region above the bump). more ophiticated (and probably numerical) method would be required to compute the potential in uch a cae.
4 Lecture 15-03-06-017.nb Theory Current Current i meaured in ampere (1 = 1 C ). The current denity J meaure current per unit area, J = n q u (5) where n i the denity, q i the charge, and u i the velocity of the moving charge. The current flowing along a urface S i given by I = S J da (6) For a time-independent current ditribution, the charge going into any urface mut equal the charge flowing out of the urface (if the charge i accumulating, the ituation i not time-independent). In uch a cae, we can write Circuit J = 0 (teady tate current flow) (7) Circuit can be analyzed uing the following three rule (known a Kirchhoff rule): 1. Ohm Law: The current through each element mut equal the voltage acro that element divided by the reitance of the element.. Charge conervation: The um of current going into a junction equal the um of current going out of that junction. Thi i charge conervation, J = 0, in circuit language. 3. Cloed loop in electrotatic field: The um of the potential difference taken in order around a loop of the network, a path beginning and ending at the ame node, i zero. Thi i E d = 0 for a tatic electric field in circuit language. Current Problem Synchrotron Current In a 6 gigaelectron-volt (1 GeV = 10 9 ev) electron ynchrotron, electron travel around the machine in an approximately circular path 40 meter long. It i normal to have about 10 11 electron circling on thi path during a cycle of acceleration. The peed of the electron i practically that of light. What i the current? The number of round trip that each electron make per econd i 3 108 m 40 m = 1.5 106 1, o that the charge paing any given point equal 1.5 10 6 1 N e = 1.5 106 1 1011 1.6 10-19 C = 0.0 (8) The point of thi imple problem i to demontrate that nothing in our definition of current a rate of tranport require the velocitie of the charge carrier to be non-relativitic and that there i no rule againt a given charged particle getting counted many time during a econd a part of the current. Combining the Current Denitie
Lecture 15-03-06-017.nb 5 We have 5 10 16 doubly charged poitive ion per m 3, all moving wet with a peed of 10 5 m. In the ame region there are 10 17 electron per m 3 moving northeat with a peed of 10 6 m. What are the magnitude and direction of J? Current denity i given by J = q n u, o the magnitude of the two J are J ion = ( e) n ion u ion = 1.6 10-19 C 5 10 16 1 10 5 m m 3 J elec = (e) n elec u elec = 1.6 10-19 C 10 17 1 10 6 m m 3 = 1.6 103 = 1.6 104 m (9) m (10) J ion point wet and J elec point outhwet (due to the negative charge of the electron). Therefore, the component of the total current denity equal J tot = J ion + J elec = -J ion - J elec, - J elec = -1.9 10 4, -1.13 10 4 The flipped direction of J elec i eay to remember if you tick in the actual negative charge of the electron in the current denity equation, J elec = (-e) n elec u elec. Reitor What i the equivalent reitance for adding two reitor (with reitance R 1 and R ) in erie. In a erie circuit, the current through each of the component i the ame, and the voltage acro the circuit i the um of the voltage acro each component. For a current I, the voltage drop acro each individual reitor equal I R 1 and I R, and therefore the total voltage drop i I (R 1 + R ). Therefore, thee two reitor are identical to a ingle reitor with reitance R = R 1 + R. By olving the loop equation for the circuit hown below, derive the rule for adding reitor in parallel. (11) To obey Kirchhoff econd rule (current in equal current out), we define the current I 1 and I to flow along the loop hown. The current through the middle egment equal I 1 - I to the right. If we ultimately end up with a negative value, it mean that we choe the wrong direction for one of our current, but that i perfectly fine (jut like finding a negative force actually mean that the force point in the oppoite direction). We now want to ue Kirchhoff third rule, which tate the voltage drop acro any loop i 0. To do thi, we label different point on our circuit a, B, C, D, E, and F, a hown below. We now pick a direction (we will chooe
6 Lecture 15-03-06-017.nb point pick (we clockwie for each loop) and calculate the total voltage drop around each loop. B C D E F For the top loop tarting at point we move clockwie to point B. The voltage drop (i.e. the difference between the voltage V B at point B and the voltage V at point ) i -I R acro thi reitor. We then continue clockwie until point D (where the voltage i the ame a point B) and then to point C. The voltage drop from D to C due to the current I i -I R 1, while the voltage drop due to the current I 1 i +I 1 R 1 (note the plu ign! For reitor there i a voltage drop in the direction of the current, o for I 1 the voltage at D i lower than the voltage at C, but ince we are looking at the top loop we are moving in the revere direction and computing the voltage at C minu the voltage at D). Therefore, Kirchhoff third rule for the top loop i -(I - I 1 ) R 1 - I R = 0 (1) Similarly, moving clockwie around the bottom loop, the voltage drop from C to D i (I - I 1 ) R 1 (i.e. the revere of the voltage drop we jut computed from D to C) plu the voltage drop in moving from F to E. Since we are current i flowing from the negative plate to the poitive plate in the battery (the poitive plate alway i marked by the larger bar), the voltage drop here i the poitive value E. Therefore, Kirchhoff third rule for the bottom loop i E + (I - I 1 ) R 1 = 0 (13) Note that adding thee two loop equation yield E - I R = 0, which i the loop equation around the whole circuit. Thu we can olve for I 1, I 1 = 1 R 1 + 1 R E E R eff (14) where 1 R eff = 1 R 1 + 1 R. The current through the battery i I 1, o E = I 1 R eff tell u that the effective reitance i R eff. Once you know how to add a circuit element (uch a a reitor) in erie and parallel, you can implify any circuit with thee element down. For example, three reitor R 1, R, R 3 in parallel would have reitance 1 R eff = 1 R 1 + 1 R + 1 R 3 (15) or equivalently R eff = R 1 R R 3 R 1 R +R R 3 +R 3 R 1 (16) Some wie word from Wikipedia: an example, conider a very imple circuit coniting of four light bulb and one 6 V battery. If a wire join the battery to one bulb, to the next bulb, to the next bulb, to the next bulb, then back to the battery, in one continuou loop, the bulb are aid to be in erie. If each bulb i wired to the battery in a eparate loop, the bulb are aid to be in parallel. If the four light bulb are connected in erie, there i ame current through all of them, and the voltage drop i 1.5 V acro each bulb, which may not be ufficient to make them glow. If the light bulb are connected in parallel, the current through the light bulb combine to form the current in the battery, while the voltage drop i 6 V
Lecture 15-03-06-017.nb 7 through light battery, voltage drop acro each bulb and they all glow. In a erie circuit, every device mut function for the circuit to be complete. One bulb burning out in a erie circuit break the circuit. In parallel circuit, each light ha it own circuit, o all but one light could be burned out, and the lat one will till function.