How do you solve both of these problems? 3) Consider a capacitor composed of 1 c

Question

How do you solve both of these problems?

3) Consider a capacitor composed of 1 cm2 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. 4) 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 4 tR2 Q/ao, what is Q/43 R2 in C/m2 magnitude units. In electrons/m2?

Explanation / Answer

Here,

area ,A = 1 cm^2

A = 1 *10^-4 m^2

d = 2 *10^-4 m

V = 90 V

electric field between plates when there is vacuum = V/d

electric field between plates when there is vacuum = 90/(2 *10^-4)

electric field between plates when there is vacuum = 4.5 *10^5 V/m

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as the battery is connected ,

electric field between plates when there is mica = V/d

electric field between plates when there is mica = 90/(2 *10^-4)

electric field between plates when there is mica = 4.5 *10^5 V/m

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Capacitance = k * A * epsilon/d

Capacitance = 6 * 8.854 *10^-12 * 1 *10^-4/(2 *10^-4)

Capacitance = 2.66 *10^-11 F

4)

E = 100 V/m

as

E * 4pi * R^2 = Q/epsilon

Q/(4pi * R^2) = E * epsilon

Q/(4pi * R^2) = 100 * 8.854 *10^-12 C/m^2

Q/(4pi * R^2) = 8.854 *10^-10 C/m^2

--------------------------

Q/(4pi * R^2) = 8.854 *10^-12/(1.602 *10^-19) electrons/m^2

Q/(4pi * R^2) = 5.526 *10^7 electrons/m^2

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