To do so you will need to first find the electric

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-field in regions I, II, and III. To do so, you will need to first find the electric field due to the left plate with charge 2 Q at each position. Then due the same for the right plate with charge Q . For both cases, assume the charge distributes uniformly over the charged plate. Answer options are in the following order: E I , E II , E III . 1. Q 2 0 ( ˆ x ) , 3 Q 2 0 ˆ x, Q 2 0 ˆ x correct 2. 3 Q 2 0 ( ˆ x ) , Q 2 0 ˆ x, 3 Q 2 0 ˆ x 3. Q 0 ( ˆ x ) , 3 Q 2 0 ˆ x, Q 2 0 ( ˆ x ) 4. Q 2 0 ( ˆ x ) , 3 Q 4 0 ˆ x, Q 4 0 ( ˆ x ) Explanation: The left plate with charge 2 Q uniformly distributed on its plates give an electric field of a large uniformly charged disk or, E 2 Q I = Q 0 ˆ x, E 2 Q II = Q 0 ˆ x, E 2 Q III = Q 0 ˆ x . The right plate with charge Q uniformly distributed on its plates give an electric field of a large uniformly charged disk or, E Q I = Q 2 0 ˆ x, E Q II = Q 2 0 ˆ x, E Q III = Q 2 0 ˆ x . The fields in the three regions are given by the superposition of the E 2 Q and E Q fields: E I = Q 2 0 ˆ x, E II = 3 Q 2 0 ˆ x, E III = Q 2 0 ˆ x . 008(part2of3)10.0points Using Gauss’ law and the fields you deter- mined in Part 1, calculate the surface charge densities on the inner and outer surfaces of each plate. The answer options are presented in the form: ( σ L , σ R ) , ( σ L , σ R ). Hint: Con- sider such Gaussian surfaces as S c and S d which have one end inside a conducting plate. 1. parenleftbigg Q A , Q A parenrightbigg , parenleftbigg Q 2 A , Q 2 A parenrightbigg 2. parenleftbigg Q 2 A , 3 Q 2 A parenrightbigg , parenleftbigg 3 Q 2 A , Q 2 A parenrightbigg correct 3. parenleftbigg Q 2 A , 3 Q 2 A parenrightbigg , parenleftbigg 3 Q 2 A , Q 2 A parenrightbigg 4. parenleftbigg Q 2 , 3 Q A parenrightbigg , parenleftbigg 3 Q A , Q A parenrightbigg Explanation: We use the Gaussian surfaces S c and S d to illustrate the relationship between the sur- face charge density σ of a conducting surface and the E -field perpendicular to that surface. Evaluating Gauss’ law over surface S c (letting it’s cross-sectional area be a ), contintegraldisplay vector E d vector A = Q enc ǫ 0 ǫ 0 E II a = σ R a σ R = ǫ 0 E II = 3 2 Q A . Repeating the procedure using S d , one finds that σ L = σ R = 3 2 Q A . Similar Gauss’ law calculations on the outer surfaces leads to the final answer: parenleftbigg Q 2 A , 3 Q 2 A parenrightbigg , parenleftbigg 3 Q 2 A , Q 2 A parenrightbigg . 009(part3of3)10.0points Now consider a similar physical configura- tion for which different information is known, shown in Fig. B. This time, you know that
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forsythe (acf2464) – HW8-c – li – (55020) 6 the plates are equally but oppositely charged, separated by a distance d , and a potential dif- ference V exists between them. Determine the surface charge density on the inner sur- face of the right plate in terms of V , d and physical constants.
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  • Spring '08
  • Turner
  • Electrostatics, Magnetic Field, Electric charge, A. Gauss

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