U V GEOMETRY due 22916 0 D1 D2 Directions Each of you will h

U& V &GEOMETRY; due: 2-29-16 0 D1 D2 Directions Each of you will hav 2.7 3.2 3.7 4.2 4.7 5.0180 140 19.0 24.0 29.0 unique set of values to have fon with this upcoming weckend: Print your last mame in the spaces provided. One letter per box. Use the decoder to determine your own personal values for the variables shown 4.0 3.0 F, S, X C, D, E B, G, R H, N, V 26.0 48.0 18.0 34.0 3.0 18.0 29.0 4.0 l,M,T J, M, T 42). | 26.0 | 3.0 19.0 For example: WILSON A-2.7B -4., DI-34.0, D2-19.0 K, L,Y 5.0 14.0 O,P 2.7 PLEASE CIRCLE THESE VALUES IN THE GRID SHOW ALL WORK HERE or ON SEPARATE PAPER ALL ANSWERS MUCH MATCH MINE TO 2 DECIMAL PLACES For all problems assume V-0 at o. REGULAR TETRAHEDRON Point charges are placed at the vertices of a REGULAR TERTAHEDRON as shown in the diagram. The charges marked A will be treated as +(A)x10\"Cand those marked B will be treated as-B) x 10, Aand B are the decoded values above. The distance between the charges is D1 in centimeters (also a decoded value). Determine the following: 1. The electric potential energy of the configuration: The electric potential at the center of the right face of the tetrahedron (centroid Point F) V- The electric potential at the midpoint of the left edge of the tetrahedron (Point E)V- The amount of work done by the field on moving the top charge out to work, \" a. U= b. c. d.

Solution

Regular Tetrahedron:

A ->4.2*10^-5 C
B ->-4.0*10^-5 C
D1 = 0.34m
a. P.E = k(A^2 + 4AB + B^2 )/(D1)^2 = 9*10^9*10^-10(17.64 + 16 - 67.2)/0.34 = 7.785*(-11.4104) = -88.83 J
b. E.P. F = k[(2A+B)2cos(30) + Bcos^2(30)]/D1 = 9*10^9 *10^-5 [(8.8)cos(30) -4cos^2(30)]/0.34 = 13.6 *10^4 J/C
c. E.P. E = k[(A+B)2 + (A+B)cos(30)]/D1 = 9*10^9*10^-5 *(0.2)(2.866)/0.34 = 15.17*10^4 J/C
d. Work Done = k[2B + A][A]/D1 = 9*10^9 *10^-10 * [-3.8]*[4.2]/0.34 = 42.25 Joules

Cube:

A ->4.2*10^-5 C
B ->-4.0*10^-5 C
D2 = 0.24m
a. P.E. = k[(AB+AB+AB)/D2 + (A^2+A^2+A^2)/D2*sqroot(2) + AB/D2*sqroot(3) + 2AB/D2 + 3B^2/D2*sqroot(2) + AB/D2*sqroot(3) + 2AB/D2 + 2A^2/D2sqroot(2) + AB/D2sqroot(3) + AB/D2 + 2*B^2/D2*sqroot(2) + AB/D2*sqroot(3) + 2AB/D2 + B^2/D2*sqroot(2) + AB/D2 + A^2/D2*sqroot(2) + AB/D2 ]
= 9*10^9 * 10^-10 [(3AB) + (3A^2)/sqroot(2) + AB/sqroot(3) + 2AB + 3B^2/sqroot(2) + AB/sqroot(3) + 2AB+ 2A^2/sqroot(2) + AB/sqroot(3) + AB + 2*B^2/sqroot(2) + AB/sqroot(3) + 2AB + B^2/sqroot(2) + AB + A^2/sqroot(2) + AB]/D2
= 9*10^9 * 10^-10 [(12AB) + (6A^2)/sqroot(2) + 4AB/sqroot(3) + 6B^2/sqroot(2) ]/D2
= 9*10^9 * 10^-10 [-201.6 + 12.47 - 38.797 + 67.88 ]/0.24 = -600.16 J

b. E.P C = k[A+B]*4*2/D2*sqroot(3) = 9*10^9 * 10^ -5 *8*0.2/0.24*sqroot(3) = 34.64*10^4 J /C
c. E.P F = k[2[A+B]/sqroot(2) + 2[A+B]*2/sqroot(3)]/D2 = 9*10^9*10^-5[0.2][1.414 + 2.309]/0.24 = 27.9255*10^4 J/C
d. E.P E = k[(A+B)*2 + 2[A+B]2/sqroot(5) + [A+B]2/3]/D2 = 9*10^9*10^-5(0.2)[2 + 4/sqroot(5) + 2/3]/0.24 = 33.41 * 10^4 J / C