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# capa03 - Solution Derivations for Capa#3 1 Consider two...

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Solution Derivations for Capa #3 1) Consider two charged parallel planes of infinite extent as shown above. The magnitudes of the charge densities on the two planes are equal. (For each statement select T True, F False). WARNING! You have 4 tries only for this problem. QUESTION: A) If both plates are negatively charged, the electric field at a points towards the top of the page. B) If both plates are positively charged, there is no electric field at b C) If the plates are oppositely charged, there is no electric field at c ANSWER: A) False, the field points toward the bottom of the page. B) True, the component of the electric field from each plane cancels the compo- nent from the other. C) True, the electric field vector pointing away from the positive plane is can- celled by the field vector pointing toward the negative plane. Remember to answer in True or False. For this problem, FTT 2) Consider a spherical CONDUCTING shell with NO NET CHARGE, with a point charge, + Q , placed at its center. (For each statement select T True, F False). WARNING! You have 8 tries only for this problem. QUESTION: A) The electric field at e is zero. B) The electric field at a is zero. C) The inner surface of the shell carries a charge - Q . D) The electric field at c is zero. ANSWER: A) False, there is an enclosed charge, namely the point charge at the center. The overall shell has no net charge, however. 1

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B) False, it encloses the point charge. C) True. Charge must reside on a surface. The interior of the shell cannot have any electric field. Thus, a Gaussian surface through c must have no field. The only way for there to be no field through c is for the Gaussian surface to enclose no net charge. D) True, see (C). Again, you must enter True or False for this question. FFTT for this example. 3) Consider a sphere of radius R = 8 . 5 m where a charge Q = 4 . 0 μC is uniformly distributed through the volume of the sphere. What is the magnitude of the electric field at a point half way between the center of the sphere and the surface? R = Given Q = Given The volume charge density of this sphere is ρ = Q V = Q 4 3 πR 3 This also means that Q = ρV . A point halfway from the center to the surface would have radius r = R 2 . To find the volume this smaller sphere has, V = 4 3 πr 3 = 4 3 π R 2 3 = 4 3 π R 3 8 = 1 6 πR 3 . The charge enclosed will be proportional to the amount of volume the smaller sphere has compared to the larger sphere. Thus, q enclosed = ρ 1 6 πR 3 = Q 4 3 πR 3 1 6 πR 3 = 1 8 Q. 2
Now, Gauss’s law will tell us the flux through this surface which we can then relate to the enclosed charge. E · dA = E cos θ dA = E dA = 4 πr 2 E = 4 π R 2 2 E = πR 2 E. Now, E · dA = q enclosed 0 . So, πR 2 E = 1 8 Q 0 E = Q 8 πR 2 0 = kQ 2 R 2 4) An electric flux of 159 N · m 2 /C passes through a flat horizontal surface that has an area of 0 . 82 m 2 . The flux is due to a uniform electric field. What is the magnitude of the electric field if the field points 15 above the horizontal?

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