Jackson 1.9 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

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1 Jackson 1.9 Homework Problem Solution Dr. Christopher S. Bair University of Massachusetts Lowell PROBLEM: Calculate the attractive force between conuctors in the parallel plate capacitor (Problem 1.6a: area A, separate by istance ) an the parallel cyliner capacitor (Problem 1.7: raii an a 2, separate by istance ) for (a) fixe charges on each conuctor (b) fixe potential ifference between conuctors. SOLUTION: (a) Parallel Plates Because the plates are large, flat, an close, we can neglect fringe effects. If one plate has total charge +Q an the other plate has total charge -Q, then they each have a charge ensity = Q A z an = Q A z for 0 < x < A 1/2 an 0 < y < A 1/2 The total force on the positive plate ue to the other plate is the sum of the force on all of its parts, which mathematically takes the form of an integral: = x E x x Here E is not the total electric fiel, but the electric fiel ue to the negative plate Expan in Cartesian coorinates: = x, y, z E x, y, z x y z Let us fin the force on the positively charge plate place at z = by substituting in its charge ensity: = Q A 0 A 0 A E x, y, x y Because we are neglecting fringe effects (in effect we are assuming that the plates are large enough that they appear infinite when compare to the rest of the system), the symmetry of the system requires that the electric fiel can only be a function of z. The remaining integrals thus reuce own to the area of the plate, which cancels the area in the enominator.

2 =Q E= The electric fiel can be foun by placing a Gaussian pillbox aroun a section of the each plate, as is emonstrate in etail elsewhere. Because the plate is thin, both top an bottom of the Gaussian surface have electric fiel lines crossing an contribute: 2 E n = 0 E n = ±Q We must be careful to realize that the normal of the insie of the positive plate points towars -z an the normal of the insie of the negative plate points towars +z. E pos plate = Q E neg plate = Q 2 A 0 Because of the symmetry, the fiel is constant an uniform across that gap so that E=E pos plate E neg plate E= Q A 0 But for the purposes of calculating the force, we o not use the total fiel, rather we use the fiel ue to to the negative plate because it is what acts on the positive plate. =Q E neg plate = Q 2 The right plate experiences a force in the -z irection, towars the other plate an is therefore attracte to it, as we shoul expect because opposites attract. Parallel Cyliners: Because the istance between the cyliners is large compare to their raii, we can approximate them as infinitely thin wires for the purpose of calculating the fiels. Place cyliner 1 at the origin with positive charge per unit length +λ an cyliner 2 at x = with negative charge per unit length -λ. If we want to fin the force on cyliner 2, we only nee to fin the fiel ue to cyliner 1 because cyliner 2 will not exert a net force on itself. The fiel ue to cyliner 1 can be foun by placing a cylinrical Gaussian surface aroun it an taking

3 avantage of symmetry: E n a= q enc 0 l E r r = q enc 0 E r 2 r= q enc l 1 0 Recognize q enc /l as the charge per unit length λ: E= 2 r 0 r = = = E,0 l l = r x, y, z E x, y, z x y z y x E x, y, z x y z Cyliner 2 experiences a force in the -r irection, towars the origin, where cyliner 1 resies. (b) Parallel Plates: A fixe potential ifference V is maintaine across the capacitor as oppose to fixe charges. Potential is relate to electric fiel accoring to: E= or a one imensional fiel, this becomes: E= z or a uniform fiel this becomes: E= V where is the istance between the two points where the potential is measure.

4 The contribution from just one plate is E neg plate = V 2 We foun previously the relationship between the total electric fiel an the charge on one plate to be: E= Q A 0 so that: Q= A 0 E Q= A 0 V inally: =Q E neg plate = A 0 V Parallel Cyliners: The electric fiel ue to cyliner 1 was foun to be: E 1 = r 2 r 0 Let us restrict ourselves to points along the x axis. E 1 = x 2 x 0 The fiel ue to cyliner 2 is E 2 = x 2 x 0 The total fiel is: E= [ 1 x 1 x ] x The potential ifference between the two cyliners can be foun by integrating:

5 a 2 V = E x V = a 2 [ 1 1 x x ] x V = ln[ a 2 a 2 ] With >> a, we can simplify this V = 0 ln[ a 2] Solve for the line charge ensity: = V 0 ln[ a 2] l = E 1, 0 l = 0 V 2 x a 2 ]2 2 [ ln