Consider a capacitor composed of 1 cm^2 plates separated by a layer of mica 0.02

Question

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/epsailon_0, what is Q/4piR^2 in C/m^2? magnitude_______units________ In electrons/m^2?________

Explanation / Answer

Vo = Eo d

Eo = Vo /d

Eo = 90 / ( 0.02x 10-2 ) = 450000 V/m

Eo = 450000 V/m or N/C

Dielectric has an inverse effect on the Electric field. That is, more the dielectric constant, lesser will be the field

Dielectric constant K = 6

So, new Electric field E = Eo /K = 450000 /6 = 75000 V/m or N/C

If the region between the plates is vacuum, then capacitance Co = Ao /d

where o the permittivity of vacuum whose value is 8.854187817... × 1012 F·m1

Area A = 1 cm2 =0.0001 m2

distance between the plates d = 0.02 cm = 0.0002 m

therefore Co = 10-4 x 8.854187817... × 1012 / 2x10-4 = 4.427093594 x 10-12 F

Co = 4.427 pF

Dielectric constant has a direcct proportionality with the capacitance. That is more the dielectric constant, more will be the Capacitance.

So, new capacitance C = K Co = 6 x 4.427 x 10-12 = 26.562 x 10-12 Farads

So, the capacitance of the mica filled capacior C = 26.562 pF

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