11. In order to move a 7.5 C charge from negative to positive terminal of a capa
ID: 1872635 • Letter: 1
Question
11. In order to move a 7.5 C charge from negative to positive terminal of a capacitor of width 5 m , we need to apply a force of 3.0 N. What is the potential difference across the capacitor plates? A. 3.75 V B. 2.5V C. 2 V D. 22.5 V E. 15 V 12. How much work is done by a uniform 300 N/C electric filed on a charge of 6.1 mC in accelerating it thru a distance of 0.20 m? A. 3.66 x 10 J B..3.66 x 10J C. 7.1 x 10 D. 9.84J E. 0.71 13. A solid conducting sphere is given a positive charge Q. How is the charge Q distributed in or on the sphere? A. It is concentrated at the center of the sphere B. It is uniformly distributed throughout the sphere. C. Its density decreases radially outward from the center. D. Its density increases radially outward from the center. E. It is uniformly distributed on the surface of the sphere only. Part II: Show all work for partial credit. I a. Find the coulomb force between the two charges on the right. G- 0.20m:@ b. If ql is removed, Find the magnitude and direction of the electric field at point x where the ql used to be 0.20 m -4.0 uCExplanation / Answer
11. given q = 7.5 micro C
capacitor width, w = 5 micro m
force F = 3 N
potential difference = V
now electric field = E
E = V/w
also, F = qE
hence
F/q = V/w
V = Fw/q = 3*5*10^-6/7.5*10^-6 = 2 V
hence option c
12. given E = 300 N/C
q = 6.1 mC
d = 0.2 m
hence
work done = F*d
F = qE
hence
W = qEd = 0.366 J
option B
13. for a conducting sphere, all the charge resides on the surface because net electric field inside a conductor is 0 and from gauss law this means that there is no charge inside the conductor
hence
option E.
1a. couloumb force = kQ1Q2/d^2
F = 8.98*10^9*(-4*10^-6)(3*10^-6)/0.2^2 = -2.694 N ( -ve sign eans the force is attractive in nture)
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