Question

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PHYS 272 Problem Set 2, Turn-in Part Please provide one page per problem and ensure that your name is written on the top of each page. Otherwise, both problems may not be graded Problem 1 In the figure shown below, a solid sphere of radius a- 3.90 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge qi = +6.70 fC; the shell has a net charge q2 = qi. What is the magnitude of the electric field at radial distances (a)0 cm, (b) r- a/2.00, (c) r = a, (d) r = 1.50a, (e) r = 2.30a, and (f) r = 3.50a? what is the net charge on the (g) inner and (h) outer surface of the shell? Problem 2 (a)What is the magnitude of an electron's acceleration in a uniform electric field of magnitude 1.87 x 106 N/C? (b)How long would the electron take, starting from rest, to attain 1/15-th the speed of light? (c)How far would it travel in that time?

Answer 1

(1) (a) The net electric flux of solid sphere is, ds E(47r2) Apply Gauss law Gauss law states that the net electric flux through a closed surface is equal to net charge 1 times E0 E(4xr2 The charge at the center of sphere is zero. Hence, for r 0 the electric field is zero. Therefore, E = 0 N/C (b) For ra, the electric field is, 1 E 4TE a Calculate the electric field at r = a/2 as follows: 1 qr E 47T8 a3 1 q(a/2) 4TEo (6.7 fC)(a/2) (ox10 N m/c) -m/c")(67 (9x10 N-m/c)(.7x10 c)(1/2) (0.039 m)2 =1.98x10 N/C -2

(c) For r a, the electric field is, 1 q Calculate the electric field at r =a as follows: 1 4TE a N-m /c*) -(9x10 (6.7x10* c) (0.039 m) = 3.96x10 N/C (d) For arb, the electric field is, 1 E Calculate the electric field at r =1.5a cm as follows: 1 E 4 TEO 1 _ 4TE (1.5a) (6.7x1015 C (9x103 N m-C(1.5x0.039 m) =1.76 x10 N/C (e) Forb <rc the electric field is zero, because the charge does not lie inside the shell Hence, the electric field at r=2.3a cm is |E = 0 N/C (f) For rc, the electric field is 1 E = 4TE2

Calculate the electric field at r 3.5a cm as follows: 1 E 4лео (4 +я) 1 2. r 4лео 24! 4ле, (3.5а)* (9x10 N m2/C3,5x0.039 m) -(9x10° N -m2/c°) 2(6.7x10+ C) 6.47 x103 N/C (g) The net charge on the inner surface of the shell is, =-6.7x10-15 C (h) The net charge on the outer surface of the shell is, 2q1 2(6.7x10-5 C) =13.4x1015 C

(2) (a) The electric force is, F qE Calculate the acceleration as follows; 0 N/C) (1.6x10 )(1.87x10 9.1x10 31 0.329 x1018 m/s2 -19 _ (b) Use the kinematic equation v =u+ at to calculate the time Calculate the time as follows: V-u _ - 0 15 0.329x1018 m/s2 3x108 m/s 15 0.329x1018 m/s -10 S = 0.6079x101 (c) Calculate the distance as follows: 1 S ut at 2 at2 1 2 at 2 =_ 2 )(0,6079 x10"10 s) 0.329x1018 m/s2 S =0.000608 m