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ch24-p014
Course: PHY 248, Fall 2010School: ITT Tech Pittsburgh
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The 14. charge is q = 4 0 RV = (10m) (1.0V) 10 8.99 N m 9 2 /C 2 = 1.1109 C.
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The 14. charge is
q = 4 0 RV =
(10m) (1.0V) 10 8.99 N m
9
2
/C
2
= 1.1109 C.
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15. A charge 5q is a distance 2d from P, a charge 5q is a distance d from P, and two charges +5q are each a distance d from P, so the electric potential at P isq (8.99 109 N m 2 C2 )(5.00 1015 C) 1 1 1 1 V= + = = 4 0 2d d d d 8 0 d 2(4.00 102 m) q= 5.62
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17. First, we observe that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative for x > d. Now we consider the two remaining regions on the x axis: x < 0 and 0 < x < d. (a) For 0 < x < d we have d1 = x and d2 = d x. LetV ( x) = kFG
Duquesne - PHYS - 264
16. Since according to the problem statement there is a point in between the two charges on the x axis where the net electric field is zero, the fields at that point due to q1 and q2 must be directed opposite to each other. This means that q1 and q2 must
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18. In applying Eq. 24-27, we are assuming V 0 as r . All corner particles are equidistant from the center, and since their total charge is 2q1 3q1+ 2 q1 q1 = 0, then their contribution to Eq. 24-27 vanishes. The net potential is due, then, to the two +4q
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19. (a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius R of the drop are related by V = q/40R. Thus8.99 10 9 N m 2 / C 2 30 10 12 C q R= = = 5.4 10 4 m. 4 0V 500 V (b) After the drops combine the total volum
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20. When the charge q2 is infinitely far away, the potential at the origin is due only to the charge q1 : q1 = 5.76 107 V. V1 = 4 0 d Thus, q1/d = 6.41 1017 C/m. Next, we note that when q2 is located at x = 0.080 m, the net potential vanishes (V1 + V2 = 0
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21. We use Eq. 24-20: V=9 2 2 30 p ( 8.99 10 N m C ) (1.47 3.34 10 C m ) = = 1.63 105 V. 2 2 9 4 0 r ( 52.0 10 m )1
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22. From Eq. 24-30 and Eq. 24-14, we have (for i = 0) p cos p cos i Wa = qV = e 2 4 0 r 2 4 0 r ep cos cos 1) = 2( 4 0 rwith r = 20 109 m. For = 180 the graph indicates Wa = 4.0 1030 J, from which we can determine p. The magnitude of the dipole moment i
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25. (a) From Eq. 24-35, we find the potential to be V =2 L / 2 + ( L2 / 4) + d 2 ln 4 0 d (0.06 m / 2) + (0.06 m) 2 / 4 + (0.08 m) 2 = 2(8.99 109 N m 2 C2 )(3.68 1012 C/m) ln 0.08 m = 2.43102 V. (b) The potential at P is V = 0 due to superposition.
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