Two circular plates of radius 6.22 cm possess a charge separation of 4.552 ?C. C

Question

Two circular plates of radius 6.22 cm possess a charge separation of 4.552 ?C. Calculate the electric field between the two plates (In N/C). Assume that the separation distance is small in comparison to the diameter of the plates. The plates are slowly pulled apart, doubling the separation distance. What changes occur with the electric field between the plates? (Again, assume the separation distance remains small in comparison to the diameter of the plates.) A) The Electric Field increases by a factor of 4 B).............................increases by a factor of 2 C) Stays the same D).............................decreases by a factor of 4 E)..............................decreases by a factor of 2

Explanation / Answer

a.E = q/e0A = 4.552 *10^-6/8.87*10^-12 * 3.14 * (6.22*10^-2)^2 = 4.22*10^9 N/C b. SINCE E is independent of d, E stays same option C PS" 10 power factor may diifer as it is not clear in the question

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