148 CHAPTER 4 Pv(2) = 2b = 10. Hence. b = 5. Appling Gauss's law to a spherical surface of radius R, J;; 1D.ds = Jv r pv d'll, DR 4nR2 = Jo rr 5R 41tR2 dr = 201t 4'" ~ ' 5 DR = 4 R2 (C/m2), A ~ 5? ') (Clm-). 4 D = RDR = R - R- Section 4-5: Electric Potential Problem 4.27 A square in the - plane in free space has a point charge of +Q at comer ai2,ai2) and the same at comer (ai2, -ai2) and a point charge of -Q at each of the other two comers. (a) Find the electric potential at an point P along the -ais. (b) Evaluate V at = all. Solution: RI = R2 and R3 = R4. With At=aI2, Rl = /(- ~r+ Ci/, R3 =,/ (+ ~)2 + (i)2. Rl = :: 2 ' a/c R~.:> -_ v) --- 2
149 a/2 P(, 0) -a/2 Q Figure P4.27: Potential due to four point charges. V = 21t o Q (2;; - v's 2) a = O.55Q 1t oa. (Problem 4.2~ The circular disk of radius a shown in Fig. 4-7 (P4.28) has unifonn charge densit Ps across its surface. (a) Obtain an epression for the electric potential V at a point P(O,O,z) on the Z-alS. (b) Use our result to find E and then evaluate it for z = h. Compare our final epression with Eq. (4.24), which was obtained on the basis of Coulomb's law. Solution: (a) Consider a ring of charge at a radial distance r. The charge contained in width dr is dq = ps(21trdr) = 21tpsrdr. The potential at P is The potential due to the entire disk is 10 Ps 1a rdr o.leo 0 r- T.:.. - V = dv = ::;- ( ", -")1/2 dv = ~ = 21tpsrdr. 41tEQR 41tEo(r2 +Z2)1/2
150 CHAPTER 4 fe p(o,a,h) h Figure P4.28: Circular disk of charge. (b) E = - r7 V = -A av a - A av () - za av az = z~ 2 0 Ps [ 1- va2 z] + Z2 The epression for E reduces to Eq. (4.24) when z = h. ~o~lem 4~ A circular ring of charge of radius a lies in the - plane and is centered at the origin. If the ring is in air and carries a uniform densit PI, (a) show that the electrical potential at (O,O,z) is given b V = PIaf[2 o(a2 +r)1/2], and (b) find the corresponding electric field E. Solution: (a) For the ring of charge shown in Fig. P4.29, using Eq. (3.67) in Eq. (4.48c) gives V(R)---1 -dl--!r PI, 1 121t--==========ad<j).. - 411: 0 II R' - 41t o $'=0 va2 +r2-2arcos PI ( ' - ) + Z2, Point (O,O,z) in Cartesian coordinates corresponds to (r,,z) = (O,,z) in clindrical coordinates. Hence, for r = 0, 1 127t PI, PIa V(O.O.z) = --.,., ad = --. /~2=..l..=7~2... 41t o 61:O..ja~+ z- 2EQva I ~
151 z z Figure P4.29: Ring of charge. (b) From Eq. (4.51), APlaa 2 _z-1/2_apla Z E = -VV = -Z2a(a Eo Z +<. ) - Z2c~( '-U a-? -; I z-'1)3/2 r:v1m). ~ Show that the electric potential difference V12 between two points in air at radial distances rj and rz from an infinite line of charge with densit pz along the z-ais is V12 = (pt/21teo) In(r2/rJ). Solution: From Eq. (4.33), the electric field due to an infinite line of charge is Hence, the potential difference is Vj2 = - E dl=- -- f:dr= it! lr! --In -.. 1'2 '2 21t or rpz 21t o PI (rz) rj Problem 4.31 Find the electric potential Vat a location a distance b from the origin in the - plane due to a line charge with charge densit PI and of length l. The line charge is coincident with the z-ais and etends from z = -l/2 to z = l/2.
155 A z=2m B z :::;-4rn Figure P4.34: Potential between B and A. GOblem~ An infinitel long line of charge with uniform densit PI = 6 (nc/m) lies 10 the - plane parallel to the -ais at = 2 m. Find the potential VAB at point A(3 m,0,4 m) in Cartesian coordinates with respect to point B(O,O,O) b appling the result of Problem 4.30. Solution: According to Problem 4.30, where rl and r2 are the distances of A and B. In this case, rl =.)(3-2?+42 = mm, r2 = 2 m. H~nce, AB = -;:------ V 2n 68.85 X 10-9 10-12 In ( V17 2 ) = -78.06 V.
-== - - 156 CHAPTER 4 z 4m A(3, 0,4).,.,., Figure P4.35: Line of charge parallel to -ais. Problem 4.36 The - plane contains a uniform sheet of charge with PSI = 0.2 (nclm- and a second sheet with PS2 = -0.2 (nc/m2) occupies the plane z = 6 m. Find VAB, VBC, and VAC for A(0,0,6 m), B(O,O,O), and ceo, -2 m,2 m). Solution: We start b finding the E field in the region between the plates. For an point above the - plane, El due to the charge on - plane is, from Eq. (4.25), El-Z-. - ~PSI 2 0 In the region below the top plate, E would point downwards for positive PS2 on the top plate. In this case, PS2 = -PSI' Hence, E = E IT'E 2=Z--Z-=Z--=Z-. ~PSI A PS2 A 2PSl A PSI 2Ea 2Eo 2Ea Ea Since E is along z, onl change in position along z can result in change in voltage. VAB = -l6 z~.zdz = _PSI zl6 = _ 6Psi _ 6 0.2 X10-9 o Eo Ea 10 Eo - 8.85 X 10-12 = -135.59 V.
157 Figure P4.36: Two parallel planes of charge. The voltage at C depends onl on the z-coordinate the lowest potential and B at the highest potential, of C. Hence, with point A being at -2 (-135.59) VBC=6VAB=- 3 = 45.20 V, VAc = VAB+ VBC = -135.59+45.20 = -90.39 V. Section 4-7; Conductors Problem 4.37 A clindrical bar of silicon has a radius of 2 mm and a length of 5 em. If a voltage of 5 V is applied between the ends of the bar and f-1e = 0.13 (m2n s), Ph = 0.05 (m2n s), Ne = 1.5 X 1016 electrons/m3, and Nh = Ne, find (a) the conductivit of silicon, (b) the current I flowing in the bar, (c) the drift velocities tie and Uh, (d) the resistance of the bar, and (e) the power dissipated in the bar.