Presentation on theme: "Lecture 9 Applications of Gauss’s Law Conductor in electric field."— Presentation transcript:
Lecture 9 Applications of Gauss’s Law Conductor in electric field
ACT: Crossed planes Which diagram corresponds to the E-field lines for these two uniformly charged infinite sheets that intersect each other as shown? +y+y +x +y+y +y+y A B C
Each sheet produces a uniform electric field. +x +y+y The total E field is uniform in each quadrant. +y+y +x
EXAMPLE: Infinite line of charge A cable of diameter D = 3 mm and length L = 200 m has a total charge Q = 4 C uniformly distributed along its length. Find the electric field at point P, located at a distance r = 2 cm from it. r P L >> r → infinite cable r >> D → one-dimensional charge distribution
Linear charge density: The elegant way (Gauss’s law) r P h The system has cylindrical symmetry. The Gaussian surface should be a cylinder of radius r and height h.
r P h E Linear charge density:
r P h
Example: Line and sheet A.0 B.2.05 × 10 5 N/C C.4.10 × 10 5 N/C D.6.15 × 10 5 N/C E.9.25 × 10 5 N/C Find the magnitude of the electric field half-way between an infinite line of uniform charge (λ = +10 μC/m) that runs parallel to an infinite sheet of uniform charge (σ = +10 μC/m 2 ). The distance between the line and the sheet is d = 1.0 m. λ = +10 μC/m σ = +10 μC/m 2 P d/2
A 2r2r Electric field produced by an infinite sheet at distance r : σ
Electric field produced by an infinite line at distance r: L r λ
λ = +10 μC/m σ = +10 μC/m 2 P d/2 E sheet E line (Answer B)
Charge in a conductor We know that E = 0 inside a conductor in equilibrium. Therefore, the electric flux though any Gaussian surface inside the conductor is zero. There is no charge inside the conductor. The charge enclosed by these surfaces is zero. The charge in a conductor in equilibrium is always on the surface(s).
Second application of Gauss’s Law: Finding the charge distribution when E is known.
Example: Conducting shell A solid non-conducting sphere with a total charge Q = +3 μC is surrounded by a concentric uncharged conducting spherical shell. What is the surface charge density σ in on the inner surface of the shell? σ out σ in Q
σ out σ in Q Negative charges are attracted towards the inner surface and positive charges are repelled towards the outer surface.
σ out σ in Q Draw a Gaussian surface inside the conducting shell. R in
Charge on the outer surface: Q -Q-Q Q The charge enclosed by the pink surface must be zero.
If the metal shell had a total charge of 3Q : 4Q4Q -Q-Q Q If the metal shell had a total charge of 3Q and the central charge was 100Q : 103Q -100Q 100Q The charge enclosed by the pink surface must be zero.
ACT: Conducting shell We now remove the uncharged shell. Compare the magnitude of the electric field at point P before and after the shell is removed. A. E before < E after B. E before = E after C. E before > E after σ out σ in Q P Q
σ out σ in Q P In both cases, the symmetry is the same. We will use the same Gaussian surface: A sphere that contains point P. And the enclosed charge is also the same: Q. Therefore, So the integral will look just the same:
ACT: Asymmetric conducting shell This time keep the shell, but move the internal charge off center. Compare the electric field at point P when the charge is centered/off center. A. E centered < E off-center B. E centered = E off-center C. E centered > E off-center Q P Q P
Q Q Charge distribution on inner surface: Uniform Concentrated near the sphere But E = 0 inside the shell in both cases! Outer surface charge is uniform (Why wouldn’t it?) E is the same for both (that of a sphere with uniform charge Q) Flux through pink surface = 0 -Q -Q-Q Total charge on inner surface is always -Q. Q Q Total charge on outer surface is always Q.
ACT: Parallel charged planes II An uncharged metal slab is inserted between the two planes as shown below. Charge densities σ L and σ R appear on the sides of the slab. Compare the electric field at point P before and after the slab is introduced. A. E before E after P σLσL σRσR σ L + σ R = 0 and distance from the plane does not matter! σ 2 = -2 μC/m 2 σ 1 = +4 μC/m 2
P σLσL σRσR How can we determine σ L and σ R ? σ1σ1 σ2σ2 The slab surfaces are two new charged planes. There are 4 contributions to the field inside the slab Note: σ 2 < 0, so that contribution really points in the opposite direction, but it’s easier to leave signs for the end.
P σLσL σRσR σ1σ1 σ2σ2 The net field inside the conducting slab is zero: