How do you solve these? Consider a capacitor composed of 1 cm^2 plates separated
ID: 1439523 • Letter: H
Question
How do you solve these?
Consider a capacitor composed of 1 cm^2 plates separated by a layer of mica 0.02 cm thick. You connect the capacitors to a 90 V battery. What would the electric field in the region of space between the plates be if the plates were separated by vacuum? magnitude______units______ Suppose the dielectric constant for mica to be 6. What then is the electric field then between the plates?______ magnitude______units_______ What is the capacitance of this mica filled capacitor? Magnitude______units_______ Now consider the global capacitance, a sphere of 4000 km radius, at a distance 50 km from ground potential. If the electric field in the atmosphere is 100 Volts/m, pointing downward, and E 4piR^2 = Q/epsilon_0, what is Q/4piR^2 in C/m^2? magnitude ______units ______ In electrons/m^2?________.Explanation / Answer
3) A =1cm^2, d =0.02 cm , V =90 V
Eo = V/d = 90/0.02*0.01 = 450000 V/m
E = Eo/k = 450000/6 =75000 V/m
C = Akeo/d = (1*10^-4*6*8.85*10^-12)/(0.02*0.01)
C =2.66*10^-11 F
4) R =4000 km , E =100 V/m
E.4pi*R^2 = Q/eo
Q/4piR^2 = E.eo = 100*8.85*10^-12
Q/4pi*r^2 = 8.85*10^-10 c/m^2
e =1.6*10^-19 C
Q/4pi*r^2 = 8.85*10^-10 (electrons/1.6*10^-19)/m^2
Q/4i*R^2 = 5.53*10^9 electrons/m^2
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.