20115ee1_1_Hw2 solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

### Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online