HW1 questions+solutions PDF

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Homework 1 questions and solutions written by instructor Petar Maksimovic....

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Homework 1 Solutions Carrie Filion Physics 104 February 9, 2019 Exercise 1. A washer made of nonconducting material lies in the x - y plane, with the center at the coordinate origin. The washer has an inner radius a and an outer radius b (so it looks like a disk of radius b with a concentric circular cut-out of radius a). The surface of the washer is uniformly charged with a surface charge density σ > 0. Solution 1. a) Here we want to find the electric R field along the vertical axis from all of the little parts of the washer, so we use the E = kσdA . We integrate over σdA to sum up all R2 of the charge that is on the surface - once this is summed up, we have our total charge. The symmetry of this problem makes cylindrical coordinates a good choice, which means that dA = rdrdθ. Note that r denotes the radial direction, and R is the distance between z and a point on the washer (i.e. the hypotenuse of a triangle drawn where z is the vertical distance and r is the radial distance of a point on the ring). The theta direction just goes from zero to 2π, so after integration it contributes a 2π. If you haven’t taken calculus to the level of knowing what a Jacobian is, you can also get the same result by knowing that the area is of a circle, and thus you have 2πrdr. Since this is a uniform charge density, we know that R b kσrdr only the vertical charge survives, so we now have Ez = 2π a R2 cosθ. Using trigonometry, √ we know that R = z 2 + r 2 and cosθ = Rz = √z 2z+r2 . Plugging this in, we see that Ez = Rb . In order to compute this integral, we need to perform u-substitution. We 2π a (z 2kσrzdr +r2 )3/2 2 set u = r + z 2 , and thus du = 2rdr. The bounds must then change to a2 + z 2 and b2 + z 2 R b2 +z 2 to reflect the substitution. Putting this all together, we have Ez = 2πkωz a2 +z 2 2udu3/2 , which 2 2 is a much simpler integral to carry out. Computing this, we get πkσz[√−2 ]b +z , and finally u a2 +z 2 that Ez = 2πkσz[√a2 1+z 2 − √b21+z 2 ], or 2ǫσz0 [ √a21+z 2 − √b21+z 2 ]. Answers in terms of k or epsilon are equivalent - your choice! b) If we take the limit that b → ∞, we see that √b21+z 2 tends to zero. Thus, in the limit of infinite b, Ez = 2πkσz[√a21+z 2 ]. c) In the limit that b → ∞ AND a → 0, we see that Ez goes to 2πkσ, or 2ǫσ0 . z d) In the limit that a → 0 and b is finite, we see that Ez tends to 2πkσ[1 − √b2 +z 2 ]. e) If we subtract our results from part d from part c, we have 2πkσ − 2πkσ[1 − √b2 z+z 2 ] = 2πkσz √ . This is the electric field for a sheet with a hole of radius b in it, whereas in b) the b2 +z 2 hole is of radius a. If we just set a = b = R, it becomes immediately evident that these are the same expressions. 1

f ) In order to perform the Taylor expansion, we must first rewrite √b2z+z 2 to be in the form of (1 + x)α. This can be done by writing √b2z+z 2 as √ z 2 2 , and identifying x as z

b2 z2

− 21 .

1+b /z

2

and α as Then, we see that (1 + x) ≈ 1 + αx ≈ 1 − 12 bz 2 . We can do this 2 Taylor expansion as we know that zb2 is quite small. Plugging this expansion in, we see 2 2 , which is the electric field for a single charge at a that Ez ≈ 2πkσ(1 − (1 − −21 bz 2 )) ≈ πkσb z2 distance z 2 . This is exactly as we would anticipate - far enough away, it should look like a point source. α

Exercise 2. A thin ring of radius a in the x - y plane is centered at the coordinate origin, and is charged with linear charge density which depends2 on the polar angle θ as λ(θ) = λ0 sinθ, where λ0 > 0. Solution 2. a) The function λ(θ) is simply a sine wave with an amplitude of λ0 . b) The symmetry of this system becomes evident when thinking of the unit circle. The top half of the unit circle has positive sine values, and the bottom half has negative values of the same magnitude. This tells us that the top half of our ring has positive charge, and the bottom half has negative charge of the same magnitude. Since field lines point from positive to negative, we know that the field lines go from north to south, pointing in the negative y direction. The field must get stronger as one approaches the y-axis, and weaker as one approaches the x-axis, as sin(π) = 1 and sin(0) = 0. In the unit circle, there is also left and right symmetry - if you fold the circle across the y-axis, the sine values that line up are equal (ex sin(π/6) = sin(5π/6)). This tells us charges in the x direction cancel, as they are equal and on opposite sides of the y-axis. Thus, our drawing will have field lines pointing in the negative y direction, increasing in density towards the y-axis. c) From our symmetry arguments above, we know that only Ey contributes to the total field, but I will also calculate Ex to show that it does not contribute. Note that we again use cylindrical coordinates, and that r is negative, as it is pointing from the ring to the origin. Since this only have rdθ, where r comes from the Jacobian. R is a line charge, R 2π we kλ0 sinθcosθdθ Ex = − kλcosθrdθ = kλ0 r1 [ 12 cos2 θ]02π = 0 =− 0 r2 R kλsinθrdθ R 2π kλ0 sinθr2 dθ 0 Ey = − = −kλo r1 [ 12 (θ − sinθcosθ)]02π = −πkλ =− 0 . r r2 r Since there is no x component, the entirety of the electric field points in the negative y 0 direction with magnitude −πkλ . This means that the electric field is oriented at negative r ninety degrees from the x-axis (positive 90 degrees would be pointing in positive y direction).

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