Homework 4 PHYS 212 Dr. Amir

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1 Homework 4 PHYS Dr. Amir. (I) A uniform electric field of magnitude 5.8 passes through a circle of radius 3 cm. What is the electric flux through the circle when its face is (a) perpendicular to the field lines, (b) at 45 to the field lines, and (c) parallel to the field lines? The electric flux of a uniform field is given by q. -b. (a) ( ) ( ) Φ = A r g r = Acosθ = 58 N C π.3m cos = 3Ng m C (b) ( ) ( ) Φ = A r g r = Acosθ = 58 N C π.3m cos 45 = Ng m C (c) Φ = A r g r = Acosθ = ( 58 N C) π(.3m) cos9 = 6. (I) Figure 6 shows five closed surfaces that surround various charges in a plane, as indicated. Determine the electric flux through each surface, S, S, S3, S 4, and S. The surfaces 5 are flat pillbox surfaces that extend only slightly above and below the plane in which the charges lie. The net flux through each closed surface is determined by the net charge inside. Refer to the picture in the textbook. ( ) ( ) Φ = 3 = ; Φ = 3 = ; ( ) Φ = 3 = ; Φ = ; Φ = (II) Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities Ignore edge effects and use Gauss s law to show (a) that for points far from the edges, the electric field between the plates is = and (b) that outside the plates on either

2 Homework 4 PHYS Dr. Amir side the field is zero. (c) How would your results be altered if the two plates were nonconductors? (See Fig. 3). Since the charges are of opposite sign, and since the charges are free to move since they are on conductors, the charges will attract each other and move to the inside or facing edges of the plates. There will be no charge on the outside edges of the plates. And there cannot be charge in the plates themselves, since they are conductors. All of the charge must reside on surfaces. Due to the symmetry of the problem, all field lines must be perpendicular to the plates, as discussed in xample -7. (a) To find the field between the plates, we choose a gaussian cylinder, perpendicular to the plates, with area A for the ends of the cylinder. We place one end inside the left plate (where the field must be zero), and the other end between the plates. No flux passes through the curved surface of the cylinder. r r r r r r r r g da g da g da g da encl = = = ends side right end σa σ A= = between between The field lines between the plates leave the inside surface of the left plate, and terminate on the inside surface of the right plate. A similar derivation could have been done with the right end of the cylinder inside of the right plate, and the left end of the cylinder in the space between the plates. (b) If we now put the cylinder from above so that the right end is inside the conducting material, and the left end is to the left of the left plate, the only possible location for flux is through the left end of the cylinder. Note that there is NO charge enclosed by the Gaussian cylinder. r r r r r r r r gda gda gda gda encl = = = ends side left end A= = outside outside (c) If the two plates were nonconductors, the results would not change. The charge would be r outside σ σ r between σ σ

3 Homework 4 PHYS Dr. Amir distributed over the two plates in a different fashion, and the field inside of the plates would not be zero, but the charge in the empty regions of space would be the same as when the plates are conductors. 7. (II) Two thin concentric spherical shells of radii r and r ( ) r<r contain uniform surface charge densities and respectively (see Fig. 3). Determine the electric field for (a) <r<r, (b) r<r<r, and (c) r>r. (d) Under what conditions will = for r>r? (e) Under what conditions will = for r<r<r? Neglect the thickness of the shells. (a) In the region < r < r, a gaussian surface would enclose no charge. Thus, due to the spherical symmetry, we have the following. encl r gd Ar = ( 4π r ) = = = (b) In the region r < r < r only the charge on the inner shell will be enclosed., σ 4πr σ r r gd Ar = ( 4 π r ) = = = encl r (c) In the region r < r, the charge on both shells will be enclosed. σ 4πr σ 4πr σ r σ r r gd Ar = ( 4 π r ) = = = (d) To make = for r < r, we must have σ opposite charge. encl r r σ r. = This implies that the shells are of (e) To make = for r < r < r, we must have σ =. Or, if a charge = 4πσ r were placed at the center of the shells, that would also make =. 35. (II) A thin cylindrical shell of radius R is surrounded by a second concentric cylindrical shell of radius R (Fig. 35). The inner shell has a total charge and the outer shell Assuming

4 Homework 4 PHYS Dr. Amir the length, of the shells is much greater than R or R, determine the electric field as a function of R (the perpendicular distance from the common axis of the cylinders) for (a) <R<R, (b) R<R<R, and (c) R>R. (d) What is the kinetic energy of an electron if it moves between (and concentric with) the shells in a circular orbit of radius ( ) thickness of shells. R R / Neglect? The geometry of this problem is similar to Problem 33, and so we use the same development, following xample -6. See the solution of Problem 33 for details. We choose the gaussian cylinder to be the same length as the cylindrical shells. encl encl r d A r g = ( π ) = = π encl (a) For < R< R, no charge is enclosed, and so = =. π (b) For R < R< R charge is enclosed, and so =,, radially outward. π (c) For R > R, both charges of and are enclosed, and so encl = =. π (d) The force on an electron between the cylinders points in the direction opposite to the electric field, and so the force is inward. The electric force produces the centripetal acceleration for the electron to move in the circular orbit. e v e = = = = = centrip F e m K mv π R 4π l Note that this is independent of the actual value of the radius, as long as R < R< R.

5 Homework 4 PHYS Dr. Amir 46. (III) A flat slab of nonconducting material (Fig. 4) carries a uniform charge per unit volume, r. The slab has thickness d which is small compared to the height and breadth of the slab. Determine the electric field as a function of x (a) inside the slab and (b) outside the slab (at distances much less than the slab s height or breadth). Take the origin at the center of the slab. Because the slab is very large, and we are considering only distances from the slab much less than its height or breadth, the symmetry of the slab results in the field being perpendicular to the slab, with a constant magnitude for a constant distance from the center. We assume that ρ > and so the electric field points away from the center of the slab. (a) To determine the field inside the slab, choose a cylindrical gaussian surface, of length x < d and cross-sectional area A. Place it so that it is centered in the slab. There will be no flux through the curved wall of the cylinder. The electric r r field is parallel to the surface area vector on both ends, and is the same magnitude on both ends. Apply Gauss s law to find x x the electric field at a distance x < d from the center of the d slab. See the first diagram. r r r r r r r r encl ( xa) gda = gda gda = gda = A= ρ ends side ends x = ρ ; x < d inside d (b) Use a similar arrangement to determine the field outside the slab. Now let x > d. See the second diagram. r r r r encl gda = gda = ends ( da) ρd A = ρ = ; x > d outside r x d d x r

6 Homework 4 PHYS Dr. Amir Notice that electric field is continuous at the boundary of the slab.