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MT2_review_solutions

# MT2_review_solutions - Physics 7B Saturday Midterm 2 Review...

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Physics 7B: Saturday Midterm 2 Review Solutions Natania Antler (Dated: April 3, 2010) 1. CHARGED SLABS ρ + ρ - ρ 2d d d z Z = 0 FIG. 1: Charged slabs for parts (a) and (b). Consider charged slabs as shown in Figure 1 . (a) Consider a infinite plane slab in the xy plane of thickness 2d, which has a uniform charge density ρ . Find the E field everywhere as a function of z (the slab is centered about z = 0). To calculate the E-field inside the slab, choose a gaussian cylinder centered on z = 0, with its two ends of some area A, and its height as 2z. I ~ E · ~ da = | E | I da = | E | 2 A = Z ρ 0 dV = ρ 0 Z dV = ρ 0 A (2 z ) , → | E | = ρ 0 z (1) So with vectors, we have: ~ E = ( ρ 0 z z ) 0 < z < d, ρ 0 z ( - ˆ z ) - d < z < 0 Outside of the slab, we have: I ~ E · ~ da = | E | I da = | E | 2 A = Z ρ 0 dV = ρ 0 Z dV = ρ 0 A (2 d ) , → | E | = ρ 0 d (2) With: ~ E = ρ 0 d z ) z > d ρ 0 d ( - ˆ z ) z < - d (b) Now find the E field everywhere for two slabs stacked on top of each other, each with thickness d. The top slab has charge density - ρ and bottom has charge density + ρ . Here we’ll need to use the principle of superposition, and also keep in mind that each slab is half as thick as the slab calculated in (a) and also that each one has a center shifted by ± d/ 2. For the top slab we calculate for the E-field, remembering that it has charge density - ρ , and we have shifted the z coordinate as used in (a) to z z - d 2 within the slab: ~ E top = - ρd 2 0 z ) z > d - ρ 0 ( z - d 2 )(ˆ z ) 0 < z < d ρ 0 d 2 z ) - d < z < 0 ρ 0 d 2 z ) z < - d

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2 For the bottom we use the coordinate shift within the slab z z - d 2 : ~ E bottom = ρd 2 0 z ) z > d ρd 2 0 z ) 0 < z < d ρ 0 ( z + d 2 )(ˆ z ) - d < z < 0 - ρ 0 d 2 z ) z < - d Adding these two together we find: ~ E total = 0 | z | > d ρ 0 ( - ( z - d 2 ) + d 2 )(ˆ z ) = ρ 0 ( - z + d )(ˆ z ) 0 < z < d ρ 0 ( z + d )(ˆ z ) - d < z < 0 (c) Can you justify why you used the Gaussian surface(s) that you did for (a) and (b)? For (a) we used a gaussian cylinder centered about z =0, since we know by symmetry that the E-field must be zero at z = 0. In (b) we can’t use a gaussian cylinder about z = 0, since we will get a net charge enclosed of 0, and a net flux of zero, yielding 0=0 and no useful information. (This is the same as the case of a dipole enclosed by a gaussian surface. Flux through the surface is 0, as is the net charge enclosed, but the E-field is clearly not zero). Instead for (b) we use the field of each slab individually (remembering to shift their centers from z = 0 to z = ± d/ 2). 2. HOLE IN A PLANE Imagine a uniformly positively charged plate in the xy plane with a hole cut into it. The hole has diameter w, and the plane has surface charge σ . (a) What is the electric field as a function of z directly above the hole? Assume z w . Using the idea of superposition, we can see that as long as we are far enough away from the plane ( z w ) we can write the field as a superposition of a positively charged infinite plane and a negative point charge of total charge Q hole = - π ( w/ 2) 2 σ . We find for some position z directly above the hole: ~ E tot = ~ E plane + ~ E hole = σ 2 0 - w 2 σ 16 0 z 2 (3) (b) What is the potential as a function of z? Again, assume z w . (Hint 1: You will need to be careful about choosing your reference point. Hint 2: Check your answer by taking the gradient, recalling that in this case
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