Physics 2212 G Quiz #2 Solutions Spring 2018

Transcription

1 Phyic 2212 G Quiz #2 Solution Spring 2018 I. (16 point) A hollow inulating phere ha uniform volume charge denity ρ, inner radiu R, and outer radiu 3R. Find the magnitude of the electric field at a ditance 2R from the center of the phere. Expre your anwer in term of parameter defined in the problem, and phyical or mathematical contant Ue Gau Law, ϵ 0 Φ = q in. Let u firt find the electric flux. Chooe a Gauian Surface with the ymmetry of the charge ditribution and paing through the point at which the electric field will be found. Thi i a phere with radiu 2R, centered at the center of the charge ditribution. Φ = E da = E4π (2R) 2 = 16πR 2 E Next, find the charge within the Gauian Surface. Since the volume charge denity i uniform Putting thee together ρ = q [ 4 q in = ρv ol = ρ V ol 3 π (2R)3 4 ] 3 π (R)3 = 4 3 πρ ( 7R 3) ϵ 0 Φ = q in ϵ 0 16πR 2 E = 4 3 πρ ( 7R 3) E = 7ρR 12ϵ 0 Quiz #2 Solution Page 1 of 6

2 II. (16 point) An infinite olid inulating cylinder ha radiu 2R, a illutrated. It volume charge denity, ρ, varie with ditance r from the center according to ρ = ρ 0 R r where ρ 0 i a poitive contant. Find the electric field magnitude at a ditance 3R from the center in term of parameter defined in the problem, and phyical or mathematical contant. Ue Gau Law, ϵ 0 Φ E = ϵ 0 E da = qin. Chooe a urface that pae through the point at which the electric field i to be found, and with the ame ymmetry a the charge ditribution. A finite cylinder of radiu 3R and length L atifie thee condition. Note that the flux through the end of the Gauian Surface i zero, a the electric field vector are perpendicular to the outward-pointing area vector. The electric field ha contant magnitude over the curved ide of the Gauian Surface, and i parallel to the outward-pointing area vector. Φ E = E da = E co θ da = E co (0 ) da = EA = E2πrL = E2π (3R) L = E6πRL The charge inide the Gauian Surface can be found from the volume charge denity, a ρ = dq/dv. Chooe a thin cylindrical hell for the volume element, dv = 2πrL dr. q in = ρ dv = = ρ 0 R2πLr 2R 0 2R 0 ρ 0 R r 2πrL dr = ρ 0R2πL 2R dr = ρ 0 R2πL (2R 0) = ρ 0 4πR 2 L 0 So ϵ 0 Φ E = q in ϵ 0 E6πRL = ρ 0 4πR 2 L E = 2ρ 0R 3ϵ 0 1. (6 point) In the problem above, what i the direction of the electric field at a ditance 3R from the center? Since ρ 0 i poitive, ρ = ρ 0 R/r i alo poitive, and the hollow cylinder i poitively charged. Electric field vector point away from poitive charge. Away from the center. Quiz #2 Solution Page 2 of 6

3 III. (16 point) Two poitive charge +Q are fixed at the vertice of an equilateral triangle with ide of length. A particle of poitive charge +q and ma m i poitioned at the apex of the equilateral triangle a hown. It i launched from that point with an initial velocity v 0 along the center line a hown. What mut the minimum initial peed v 0 of thi particle be o that it pae between the two fixed charge? Expre your anwer in term of parameter defined in the problem and phyical or mathematical contant. (NOT on Earth no gravity!) Ue the Work-Energy Theorem, W ext = K + U + E th Chooe a ytem coniting of all three charged particle. There i no work done by external force on that ytem, and there are no non-conervative force converting mechanical energy to thermal energy within that ytem. The potential energy i that of a ytem of charged particle. So 0 = (K f K i ) + (U f U i ) = ( ) 1 2 mv2 f 1 2 mv2 i + (K Qq K Qq ) ( + K Qq K Qq ) ) + (K Q2 K Q2 r f r i r f r i r f r i where the potential energy of the ytem i the um of the potential energie aociated with each pair of particle. Note that the final peed of the particle with charge q mut be only infiniteimally more than zero a it pae between the particle with charge Q. 0 = ( mv2 0 ) + (K Qq /2 K Qq ) ( + K Qq /2 K Qq ) ) + (K Q2 K Q2 Solve for v 0 : ( 1 2 mv2 0 = 2 K Qq /2 K Qq ) + 0 = 2K Qq v 0 = 4K m Qq 2. (6 point) Conider a ituation in which the particle with charge q in the problem above were replaced by a particle with charge q = 2q, and the fixed charge Q were each replaced with fixed charge Q = 2Q. How doe the minimum peed, v 0, required for the particle to pa the fixed charge in thi ituation, compare to your anwer v 0 above? Since the electric potential energy depend on the product of the pair of charge, doubling all the charge value will increae the potential energy by a factor of 4. Initially, then, the moving particle will need four time the kinetic energy to pa between the fixed charge. A the kinetic energy depend on the quare of the peed, doubling the peed will provide four time the kinetic energy. v 0 = 2v 0 Quiz #2 Solution Page 3 of 6

4 3. (8 point) Three particle, each with charge Q, are located a hown on different corner of a rhombu with ide of length a and a diagonal of length a (a rhombu ha 4 equal length ide that do not interect at right angle). With repect to zero at infinity, what i the electric potential at the empty vertex? Since electric potential i a calar, the potential at the empty vertex i jut the um of the potential contributed by each of the three charge Q. With repect to zero at infinity, the electric potential due to a ingle point charge i V = Kq/r. Each of the charge Q at the end of the diagonal with length a are a ditance a from the empty vertex, and o each contribute electric potential KQ/a. The remaining charge i a ditance a co (30 ) = a 3/2 from the diagonal of length a, and o i twice that ditance, or a 3 from the empty vertex. Therefore, it contribute electric potential KQ/a 3. The total potential at the empty vertex i the um K Q a + K Q a + K Q ( a 3 = ) K Q 3 a 4. (8 point) An infinite lab with thickne 2h ha uniform volume charge denity ρ. The lab i infinite in the x and y direction and centered at the origin, extending from h to +h along the z axi. A finite egment of the lab i illutrated. Are there any location where the magnitude of the electric field i zero, and if o, where? Conider the ymmetry of the lab. If it i, for example, rotated 180 about the x axi, the charge ditribution i unchanged. The electric field mut, therefore, alo be unchanged. The 180 rotation about the x axi would revere the direction of any field on the x y plane. The only way an electric field could be both revered and unchanged i if it magnitude were zero. Ye, the field ha zero magnitude only on the x y plane, z = 0. Quiz #2 Solution Page 4 of 6

5 5. (8 point) A hollow conductor, illutrated in cro-ection, carrie a net charge of -3 nc. Within it void lie a particle with a charge of +5 nc. What i the net charge on the inner and outer urface of the conductor at equilibrium? Conider a Gauian Surface within the olid part of the conductor. The field in the olid part of a conductor at equilibrium i zero, o the flux through thi Gauian Surface i zero, o the net charge contained within the Gauian Surface mut be zero. There mut be a charge of -5 nc on the inner urface of the conductor to balance the +5 nc charge on the particle. Charge i conerved. If there i -5 nc on the inner urface of the conductor, but the conductor ha a net charge of -3 nc, then there mut be +2 nc on the outer urface. Q inner = 5 nc while Q outer = +2 nc 6. (8 point) A poitive point charge +q lie at the center of a tetrahedron, contructed of four equilateral triangle with edge a. What i the electric flux through the bottom face of the tetrahedron? Each face i identical, o one-fourth of the total flux mut pa through each face. Ue Gau Law. ϵ 0 Φ = q in Φ/4 = +q/4ϵ 0 Quiz #2 Solution Page 5 of 6

6 7. (8 point) Three iolated ytem, i, ii, and iii, each conit of a negatively charged particle q and two poitively charged particle +q, all with the ame charge magnitude. Let the configuration with zero electric potential energy be the ame for each ytem. Rank the electric potential energie U of the ytem from greatet to leat. (Remember that ince the ytem are iolated, there i no interaction between them.) Recall that the electric potential energy of a ytem of two point charge i U = Kq 1 q 2 /r if infinite eparation i choen a the zero point, and that the electric potential energy of a ytem of multiple point charge i jut the um of the electric potential energie of all the pair. Compare ytem i and ytem ii. In ytem ii, the two poitively charged particle are the ame ditance apart a in ytem i, o there i no electric potential energy difference due to that interaction. The negatively charged particle i the ame ditance from one of the poitively charged particle, but it i twice a far from the other one. Work would have to be done on ytem i to convert it to ytem ii, a the negatively charged particle i attracted to the poitively charged particle i it being pulled farther from. So, U ii > U i. Next, compare ytem i and ytem iii. In ytem iii, the negatively charged particle i the ame ditance from the poitively charged particle a in ytem i, o there i no electric potential energy difference due to thoe interaction. The poitively charged particle, however, are twice a far apart. Energy would be releaed a ytem i i converted to ytem iii, a the two poitively charged particle repel each other. So, U iii > U i. U ii > U i > U iii Quiz #2 Solution Page 6 of 6