148

Pv(2) = 2b = 10.

Hence. b = 5.

Applying Gauss's law to a spherical surface of radius R,

1D.ds = r pv d'll,J;; Jv

rR ~DR ·4nR2 = Jo 5R· 41tR2 dR = 201t 4'" '5

DR = 4 R2 (C/m2),

A ~ 5? ')D = RDR = R - R- (Clm-).4

Section 4-5: Electric Potential

CHAPTER 4

Problem 4.27 A square in the x-y plane in free space has a point charge of +Q atcomer aI2,aI2) and the same at comer (aI2, -aI2) and a point charge of -Q ateach of the other two comers.

(a) Find the electric potential at any point P along the x-axis.(b) Evaluate V at x = all.

Solution: RI = R2 and R3 = R4.

With

Rl = /(x- ~r+ Ci/,

R3 = ,/ (x+ ~)2 + (i)2.

Atx=aI2,

Rl = ::2 '

a/CR~ _ v).:> - ---

2

149

xP(x, 0)

-a/2 Q

y

a/2

rdr

Q (2 2) O.55QV = 21t£o ;; - v's a = 1t£oa .

Figure P4.27: Potential due to four point charges.

10 Ps 1aV = dV = ::;- ( " , -")1/2o .Leo 0 r- T .:..-

dV = ~ = 21tpsrdr .41tEQR 41tEo(r2 +Z2)1/2

The potential due to the entire disk is

dq = ps(21trdr) = 21tpsrdr.

The potential at P is

CHAPTER 4

(Problem 4.2~ The circular disk of radius a shown in Fig. 4-7 (P4.28) has unifonncharge density Ps across its surface.

(a) Obtain an expression for the electric potential V at a point P(O,O,z) on theZ-axlS.

(b) Use your result to find E and then evaluate it for z = h. Compare your finalexpression 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

r7 A av A av A av ~ Ps [ z]E = - y V = -x ax - y ()y - z az = z 2£0 1- va2 + Z2 •

The expression for E reduces to Eq. (4.24) when z = h.

CHAPTER 4

fEp(O,a,h)

h

Figure P4.28: Circular disk of charge.

x

(b)

150

~o~lem 4~ A circular ring of charge of radius a lies in the x-y plane and iscentered at the origin. If the ring is in air and carries a uniform density PI, (a) showthat the electrical potential at (O,O,z) is given by 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

1 !r PI, 1 121t PI ,V(R)--- -dl-- --==========ad<j).

. - 411:£0 II R' - 41t£o $'=0 va2 + r2 - 2arcos (¢' - ¢) +Z2

Point (O,O,z) in Cartesian coordinates corresponds to (r,¢,z) = (O,¢,z) in cylindricalcoordinates. Hence, for r = 0,

1 127t PI , PIaV(O.O.z) = -- .,., ad¢ = --. /~2=..l..=7~2... 41t£o 61:O..ja~+z- 2EQva I ~

Problem 4.31 Find the electric potential Vat a location a distance b from the originin the x-y plane due to a line charge with charge density PI and of length l. The linecharge is coincident with the z-axis and extends from z = -l/2 to z = l/2.

151

r:v1m).

y

iT! lr! rpz PI (rz)Vj2 = - E·dl=- --·f:dr= --In - .

. 1'2 '2 21t£or 21t£o rj

Figure P4.29: Ring of charge.

z

z

(b) From Eq. (4.51),

APlaa 2 _z-1/2_APla Z

E= -VV = -Z2a(a +<. ) - Z2c~( ? I '1)3/2Eo Z '-U a- -; z-

Hence, the potential difference is

CHAPTER 4

~ Show that the electric potential difference V12 between two points inair at radial distances rj and rz from an infinite line of charge with density pz alongthe z-axis is V12 = (pt/21tEo) In(r2/rJ).

Solution: From Eq. (4.33), the electric field due to an infinite line of charge is

GOblem~ An infinitely long line of charge with uniform density PI = 6 (nC/m)lies 10 the x-y plane parallel to the y-axis at x = 2 m. Find the potential VAB at pointA(3 m,0,4 m) in Cartesian coordinates with respect to point B(O,O,O) by applyingthe result of Problem 4.30.

Solution: According to Problem 4.30,

H~nce,

155

z :::;-4rn

z=2m

B

A

rl = .)(3-2?+42 = mm,r2 = 2 m.

Figure P4.34: Potential between B and A.

V 6 X 10-9 ( 2 )AB = -;:------

2n x 8.85 x 10-12 In V17 = -78.06 V.

CHAPTER 4

where rl and r2 are the distances of A and B. In this case,

-== - -

y

CHAPTER 4

4m

z

A(3, 0,4) ••., .,.,

Figure P4.35: Line of charge parallel to y-axis.

x

In the region below the top plate, E would point downwards for positive PS2 on thetop plate. In this case, PS2 = -PSI' Hence,

E E 'E ~PSI A PS2 A 2PSl A PSI= IT 2=Z--Z-=Z--=Z-.2Ea 2Eo 2Ea Ea

Problem 4.36 The x-y 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 by finding the E field in the region between the plates. For anypoint above the x-y plane, El due to the charge on x-y plane is, from Eq. (4.25),

E - ~PSIl-Z-.2£0

156

Since E is along z, only change in position along z can result in change in voltage.

VAB = -l6 z~ .zdz = _PSI zl6 = _6Psi _ 6 x 0.2 X10-9o Eo Ea 10 Eo - 8.85 X 10-12 = -135.59 V.

Section 4-7; Conductors

The voltage at C depends only on the z-coordinate of C. Hence, with point A being atthe lowest potential and B at the highest potential,

y

157

Figure P4.36: Two parallel planes of charge.

x

CHAPTER 4

-2 (-135.59)VBC=6VAB=- 3 = 45.20 V,

VAc = VAB+ VBC = -135.59+45.20 = -90.39 V.

Problem 4.37 A cylindrical 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 conductivity 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.