EEL3472HA3S

EEL3472HA3S - H frS frJI 2o/o Problem 4.2 Find the total...

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HfrS frJI 2o/ o Problem 4.2 Find the total charge contained in a cylindrical volume defined by r 1 2 mand 0 ( z I 3 mifpu : 20rz (mC/nf). Solution: For the cylinder shown in Fig. P4.2, application of Eq. (4.5) gives e: ft ['" f 2orz r dr dQ dz Jz:o J a:o Jr=o / t2 r2n 13 :t*ro*)l I :480n(mc):1.5g. \ r l,:ola:ol":, Figure P4.2: Cylinder of Problem 4.2.

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Problem 4.14 A line of charge with uniform density fu: 8 (1tC/m) exists in air along the z-axis between z :0 aad z: 5 cm. Find E at (0,10 cm,0). Solution: Use of Eq. (a.2lc) for the line of charge shown in Fig. P4.14 gives E:-L tt4. 4neo lY" R'2 ' Rt : j}.l -?z I f'os : q;6 J,:0, 1a*rc-\rffia, _ 8x ro-6 | irc,+z ll'* 4nq Lr/(o.t), +*Jl"^ :71.86x103W4.47 -21.061 :9321.4x103 -276.2 x 103 (v/m). Figure P4.14: Line charge. p'= fo.t - 2z
Problem 4.16 A line of charge with uniform densrty p, extends between z : -Ll2 and z: L/2 along the z-axis. Apply Coulomb's law to obtain an expression for the electric field at any point P( r, O ,0) on the x-y plane. Show that your result reduces to the expression given by (a.33) as the length.L is extended to infinity. Solution: x-y plane Figure P4.16: Line charge of length I. Consider an element of charge of height dz at height z. Call it element l. The electric field at P due to this element is afE1. Similarly, an element at -z produces dE2. These two electric fields have equal z-components, but in opposite directions, and hence they will cancel. Their components along i will add. Thus, the net field due to both elements is dE : df,r * dLz - r2ficoso dz - -4;€oF- - tplcos9 dz 2nesRz

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where the cos I factor provides
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EEL3472HA3S - H frS frJI 2o/o Problem 4.2 Find the total...

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