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Lecture-7 Conductors Field At the Surface of a Conductor • E E|| E • • • • September 26, 2007 Imagine an electric field at some arbitrary angle at the surface of a conductor. There is a component perpendicular to the surface, so charges will move in this direction until they reach the surface, and then, since they cannot leave the surface, they stop. There is also a component parallel to the surface, so there will be forces on charges in this direction. Since they are free to move, they will move to nullify any parallel component of E. In a very short time, only the perpendicular component is left. Field Inside a Conductor • • • • • September 26, 2007 We can use Gauss’ Law to show that the inside of a conductor must have no net charge. Take an arbitrarily shaped conductor, and draw a gaussian surface just inside. Physically, we expect that there is no electric field inside, since otherwise the charges would move to nullify it. Since E = 0 everywhere inside, E must be zero also on the gaussian surface, hence there can be no net charge inside. Hence, all of the charge must be on the surface (as discussed in the previous slide). Field Inside a Conductor • • • • • • We can use Gauss’ Law to show that the inside of a conductor must have no net charge. Take an arbitrarily shaped conductor, and draw a gaussian surface just inside. Physically, we expect that there is no electric field inside, since otherwise the charges would move to nullify it. Since E = 0 everywhere inside, E must be zero also on the gaussian surface, hence there can be no net charge inside. Hence, all of the charge must be on the surface (as discussed in the previous slide). If we make a hole in the conductor, and surround the hole with a gaussian surface, by the same argument there is no E field through this new surface, hence there is no net charge in the hole. Field Inside a Conductor • • • if you try to deposit charge on the inside of the conductor... The charges all move to the outside and distribute themselves so that the electric field is everywhere normal to the surface. This is NOT obvious, but Gauss’ Law allows us to show this! There are two ideas here • Electric field is zero inside conductors • Because that is true, from Gauss’ Law, cavities in conductors have E = 0 September 26, 2007 A Charge Inside a Conductor Q. Spherical cavity What will happen when we add a charge inside a conductor? A. B. C. D. Conducting sphere September 26, 2007 Positive point charge E. E field is still zero in the cavity. E field is not zero in the cavity, but it is zero in the conductor. E field is zero outside the conducting sphere. E field is the same as if the conductor were not there (i.e. radial outward everywhere). E field is zero in the conductor, and negative (radially inward) outside the conducting sphere. Use Gauss’ Law to Find Out 0 E dA qenc Is E = 0 in the cavity? No, because there is charge enclosed (Gauss’ Law). Gaussian Surface Is E = 0 in the conductor? Yes, because as before, if there were an electric field in the conductor, the charges would move in response (NOT Gauss’ Law). If we enlarge the gaussian surface so that it is inside the conductor, is there any net charge enclosed? It looks like there is, but there cannot be, because Gauss’ Law says E = 0 implies qenc = 0! How do we explain this? There must be an equal and opposite charge induced on the inner surface. September 26, 2007 E Field of Charge In Conductor This negative charge acts with the inner charge to make the field radial inside the cavity. This negative charge cannot appear out of nowhere. Where does it come from? It comes from the outer surface (electrons drawn inward, attracted to the positive charge in the center). Therefore, it leaves positive charge behind. The net positive charge that appears conductor is exactly the same as the original charge in the center, so what do the field lines look like? By spherical symmetry, the positive shell of charge acts like a point charge in the center, so field is the same as the field of the original point charge. September 26, 2007 Back to the Previous Question Q. Spherical cavity What will happen when we add a charge inside a conductor? A. B. C. D. Conducting sphere September 26, 2007 Positive point charge E. E field is still zero in the cavity. E field is not zero in the cavity, but it is zero in the conductor. E field is zero outside the conducting sphere. E field is the same as if the conductor were not there (i.e. radial outward everywhere). E field is zero in the conductor, and negative (radially inward) outside the conducting sphere. E Field of Charge In Conductor What happens when we move the inner charge off-center? It induces an off-center charge distribution on the Inner wall. Note that the field lines distorted, so they remain perpendicular to the inner wall. What happens to the outer positive charge distribution? Draw a gaussian surface inside the conductor to find out. The net charge enclosed is zero, so E = 0, which we already knew because it is inside the conductor. The inner charge is shielded by the induced charge distribution, so the outer charges will be evenly distributed. September 26, 2007 Electrostatic Boundary Conditions E σ 2ε0 Discontinuity For an infinite plane carrying uniform surface charge E y Discontinuity σ 2ε0 R For a spherical shell of radius R, carrying uniform surface charge r Electrostatic Boundary Conditions From Gauss’ law S E above E below 1 σ ε0 1 1 E ds Qenc A 0 0 E above E above AE E below below 1 A A 0 1 0 Conclusion: The normal component of E is discontinuous by an amount σ/ε0 at any boundary. Electrostatic Boundary Conditions The line integral of the static electric field E around a closed path is zero E dl 0 s E || above lE E || above E || below l0 || below The tangential component of E by contrast is always continuous Electrostatic Boundary Conditions The potential is continuous across any boundary. b Vabove Vbelow E dl a As the path length shrinks to zero i.e Vabove Vbelow Basic properties • • • • • E=0 inside a conductor. =0 inside a conductor. Any net charge resides on the surface. A conductor is an equi-potential. E is perpendicular to the surface, just outside a conductor. • Note, none of this is true for insulators. Summary • Electric flux is the amount of electric field passing through a closed surface. • Flux is positive when electric field is outward, and negative when electric field is inward through the closed surface. • Gauss’ Law states that the electric flux is proportional to the net charge enclosed by the surface, and the constant of proportionality is 0. In symbols, it is q 0 enc 0 E dA qenc • There are three geometries we typically deal with: Geometry Charge Density Gaussian surface Linear l = q/L Cylindrical, with axis along line of charge E l 20 r Sheet or Plane = q/A Cylindrical, with axis along E. E 0 Spherical = q/V Spherical, with center on center of sphere E September 26, 2007 Electric field Line of Charge Conducting q 40 r 2 rR E 2 0 Nonconducting q r E 3 4 R 0 r<R Q. A metal sphere of radius R, carrying charge q is surrounded by a thick concentric metal shell. The shell carries no net charge. (a) Find the surface charge density at R, a and b (b) Find the potential at the centre, using infinity as the reference point (c) If outer surface is grounded, then find out the answer of (a) and (b) Answer: a q b (a) R R q q q , , a b 4R 2 4a 2 4b2 (b) V 0 (c) 1 q q q 4πε0 b R a 1 q q q =0 V 0 b 4πε0 b R a Surface charge on a conductor σ E above E below nˆ ε0 Electrostatic boundary condition: => Field outside a conductor σ E nˆ ε0 Force on a charge element dq placed in an external field E(e) : F dq Ee dq E On a volume charge distribution : F ρ E dτ τ On a surface charge distribution : F σ E da s “ BUT E is discontinuous across a surface charge distribution” On the surface charge the force per unit area : 1 f σ E above E below 2 Force on a conductor Force (per unit area) on the conductor surface: 1 2 f σ nˆ 2 ε0 Outward electrostatic Pressure on the conductor surface : 1 P ε0 E 2 2 Two large metal plates (each of area A) are held a distance d apart. Suppose we put a charge Q on each plate, what is the electrostatic pressure on the plates ? 2 Answer: P Q 2 ε 0 A2 Q.5 consider two potential fields V1 = y and V2 =y + ex sin y. (a) Is 2V1 = 0? (b) Is 2 V2 = 0? (c) Is V1 = 0 at y = 0? (d) Is V2 = 0 at y = 0 ? (e) Is V1 = π at y = π ? (f) Is V2 = π at y = π ? (g) Are V1 and V2 identical ? Ans: (a) yes (b) yes (c) yes (d) yes (e) yes (f) yes (g) No