PHY 2049 FALL 2000 EXAM 1

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1 PHY 09 FALL 000 EXAM 1 1. Figure below shows three arrangements of electric field lines. A proton is released from point X. It is accelerated by the electric field toward point Y. Points X and Y have eual separations for the three arrangements. Rank the arrangements according to the velocity of the proton at point Y, greatest first. Answer: ACB X Y X Y X Y A B C The charge that eperiences the largest force will achieve the greatest velocity and the largest electric field created the largest force. The electric field is greatest where the electric field lines are closest together. The field lines are most dense in A. The net is C and then B. (The diverging lines in C mean that the electric field is not uniform.). A spherical shell of radius R = 1 m is uniformly charged with a charge Q =1C.Find the pressure eperienced by the shell walls due to electrostatic repulsion of the charge. Hint: Consider a small patch on the shell surface. The net electric field on the shell surface (field of a uniformly charged shell) is the sum of the field made by the patch itself (field of uniformly charged plane) and the remaining part of the shell (E). The force on the patch is its charge times E. Pressure is the force per unit area. Answer: Pa The electric field just outside the sphere is a superposition of the electric field due to the patch and the electric field due to the remainder of the sphere. Even though the patch is small, the point is just outside the sphere and the patch appears to be an infinite plane. E sphere = E patch + E remain Q = σ + E πɛ 0 R ɛ remain 0 Q E remain = πɛ 0 R Q 1 πr ɛ 0 = kq R with k =1/πɛ 0. The force on the patch is F = (E remain )( patch ) ( ) kq = (σa) R

2 where σ =charge/area=q/πr. F A = kq Q R πr = kq 8πR P = Pa. A uadrupole is a combination of two staggered dipoles with dipole moments d that are eual but opposite in direction. Find the magnitude of the electric field made by the uadrupole at a distance d. Answer: kd / d d The electric field created by a dipole at a point located along the dipole moment points away from the positive charge and toward the negative charge and has the magnitude E = kd r where r is measured from the center of the dipole. The electric field due to the left dipole points to the left and the field due to the right dipole points to the right, E = kd ( + d/) + kd ( d/) = kd( ( + d/) +( d/) ) = kd ( 1+ d ) ( + 1 d ) kd ( ( 1+( ) d ) ( + 1+( ) d = kd = kd. ( ) d. A uniformly charged sphere of radius R with a spherical cavity of radius R/, as shown is charged with a uniform charge density ρ. Find the magnitude of the electric field at the point P at a distance R/ from the center of the sphere on the side opposite to the cavity. Answer: kπρr/ ))

3 R R/ P R/ This is a superposition of the electric fields of two bodies. E 1 is the electric field at a distance R/ inside a uniformly charged sphere of radius R and charge density ρ. E is the electric field at a distance R eterior to a sphere of radius R/ and charge density ρ. For E 1 use the electric field at a distance r from the center to a point located inside a sphere of radius R containing uniform charge (Sample Problem -7) Applied to this situation E = k R r. E 1 = k R R = k R = k (ρ Volume) R = k (ρ π ) R R = πkρr. E is found from the charge contained the the smaller sphere E = k R = k R ( ρ π ( R ) ) = πkρr. The net field is E = E 1 + E = kπρr/.. Of the eight charge distributions show in Figure which one results in the maimum magnitude of the electric field at the center point of the suare. Each configuration consists

4 of a suare with or without rods along its sides. Each rod has either +Q (filled bars) or Q (empty bars) charge. Answer: () (1) () () () () The electric field due to each bar is indicated in the diagram. If each rod creates an electric field E at the center of the suare then E 1 =E, E =0,E = E, E = E, and E = E.. A semi-infinite rod is placed along as shown in Figure and uniformly charged with positive charge density λ. Find the direction of the electric field with respect to y-ais direction at the point P 1 meter away from the tip of the rod. Hint: use integration to calculate E and E y. Answer: y P θ λ The electric field due to a small portion d is de =(kλ d)/r where r = + y.the components are de = de cos θ and de y = de sin θ, wherecosθ = /r and sin θ = y/r. Integrating E = E y = 0 0 kλ kλ d = ( + y ) / y kλy kλ d = ( + y ) / y. Since E = E y the angle is. 7. Four charges of the same sign and value are placed in the corners of a suare and free to move. One more charge Q is placed in the center of the suare so that the entire system of the five charges has become statically stable, i.e., net forces on each of the five charges are eual to zero. Find the value of the charge Q in terms of. Answer: 0.97 The charges in the corners will repel each other. In order to hold the system together, the force (F ) due to Q must be attractive. Since Q is in euilibrium for all values of Q,

5 Q F F 1 o F F we consider a corner charge. F = F 1 F cos + F cos 0 = k L k ( L) 1 + k Q ( L/) 1 = / + Q / Q = +1 Q = The sign of Q is the opposite to the sign of since their interaction must be attractive. 8. A pair of parallel horizontal conducting plates of length L, distanced apart, are charged so as to produce a uniform vertical electric field E between them. An electron (mass m and charge e) moves horizontally with velocity v and enters the electric field eactly halfway between the plates as shown in the Figure. What must be the electron s minimum velocity in order for it to pass through the field without hitting one of the plates? Ignore gravity. Answer: ee/md L L e, m v E D The horizontal position as a function of time is = vt and the vertical position is y = at /=eet /m. Solving for t = /v and substituting gives y = ee /mv. The problem specifies y = D/ when = L D = eel mv

6 v = ee md L. 9. Two positive charges Q = 1 nc are held fied 1 m apart. A test charge is inserted along the line between them and cm away from one of them. The force acting on the charge is found to be 1 mn. Find the magnitude of the test charge. Answer: C Since the two 1 nc charges have the same sign, the forces they eert on the test charge placed between them will act in opposite directions. F = F 1 F = kq kq r1 r ( r = kq r1 ) r1r = Fr1 r kq(r r1) = C. 10. The electric field in three dimensions is E =(a)î+(by)ĵ +cˆk, witha =1 N, b = N Cm and c = N. Find the electric field flu through the surface of a cube with a side of 1 m C and oriented as shown in the Figure. Answer: Nm /C Cm, y 1 z The flu through the cube is the flu through the sides of the cube Φ E = E da = E d A 1 + E d A + E d A + E d A + E d A + E da. 1

7 Recalling that da points away from the enclosed volume, da 1 = da î, da = da ĵ, da = da ˆk, da = da î, da = da ĵ, andda = da ˆk. Substituting the epression for E and the das into the epression for Φ E Φ E = a da by da cda + a da + by da + cda. 1 The first two terms are zero since = 0 on side 1 and y = 0 on side. The length of the sides of the cube are L = 1 m and the area of one side is L. The total flu is Φ E = 0 0 c da + al da + b da + c da = cl + all + bll + cl = (a + b)l = Nm /C.

Profs. D. Acosta, A. Rinzler, S. Hershfield. Exam 1 Solutions

PHY2049 Spring 2009 Profs. D. Acosta, A. Rinzler, S. Hershfield Exam 1 Solutions 1. What is the flux through the right side face of the shown cube if the electric field is given by E = 2xî + 3yĵ and the