20115ee1_1_Hw2 solutions

20115ee1_1_Hw2 solutions - 3.3a) to find the total charge...

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Unformatted text preview: 3.3a) to find the total charge we mu cm. Due to the absolute value we can c multiply by 2 b) To find electric flux leaving at & ¡ and solve 3.5) D = 4xy a x +2(x 2 +y 2 ) a y + 4yz a z we use ga ¢ £¤¥¦§¨£© ¡ ª« ∙ ¬­ ¡ ª® HW#2 Solutions ust integrate over the surface area around the infini ¢ ±§±²¦ ¡ ³ 5´ µ¶· | ¸ | ¹ ¶º µ¹ · & ∅ " change the bounds of the integral with respect to z t ¢ ±§±²¦ ¡ 2 ³ 5´ µ¶· | ¸ | ¹ ¶º · · & ∅ " Solving we get Q total =0.25 nC ¡ 8%» 1cm<z<5cm, 30 o <φ<90 o we change the boun ¡ 2 ³ 5´ µ¶· | ¸ | .·( º ¶ .·) º * & ∅ " Solving we get 9.45 pC auss’s law to integrate over the faces of the cube ®¯° + ª,¯®®¯- + ª¼/¯0® + ª,1½3 + ª4¾¼® + ite cylinder. Ρ =8 to 0 to ∞ and nds of integration + ª /7¿9® & &:¡ &=> Addi 3.6 We have an infinite slab of cha We draw our Gaussian surface as a &<@A ¢ £ B 4D"1 ¤ E ∙ B D F1 G E ¥ H ¦ ¦ Where z=5 ∮ <@A ¢ §80LM & N@<<@¨ ¢ 0 since z=0. &;[email protected]< ¢ £ B 4FD1 O E ∙ B D "1 P ( H ¦ ¦ E Where x=2. ∮ ;[email protected]< ¢ §80LM ¡;< ¢ £ 2 B F ¥ + " ¥ E 1 Q ∙ B F "BR1 Q E ( ¥ ¦ ¦ &:¡;< ¢ R§93.3333LM >?U< ¢ £ 2 B F ¥ + " ¥ E 1 Q ∙ B F "B1 Q E ( ¥ ¦ ¦ ∮ =>?U< ¢ §93.3333LM ing up all contributions Q enclosed = 360nC arge as shown a cylinder Due to the symmetry of the problem...
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This note was uploaded on 12/05/2011 for the course EL ENGR 1 taught by Professor Joshi,chand during the Fall '11 term at UCLA.

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20115ee1_1_Hw2 solutions - 3.3a) to find the total charge...

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