A parallel plate capacitor is made with two square plates of side length 8cm sep

Question

A parallel plate capacitor is made with two square plates of side length 8cm separator i by a 1.5mm air gap. What is the capacitance of this capacitor? If a positive charge of 420 mu C is placed on one side of the capacitor and a negative charge of 420 mu C is placed on one the other, what is the potential difference between the two plates? Now, consider two different possibilities: While the two sides of the capacitor remain isolated (charge is conserved), a dielectric with k = 22.8 is inserted between the two plates. What are the new values of Q, C, and V? The two plates are connected to a battery (constant potential source), a dielectric with k = 22.8 is inserted between the two plates. What are the new values of Q, C, and V?

Explanation / Answer

(a) We know that C = e0A/d
A = area of plate: 0.08m x 0.08 m = 0.0064 m^2
e0 = constant: 8.85 x 10^-12 Fm-1
d is distance between plates = 1.5 x 10^-3 (1.5mm)
C = 8.85 x 10^-12 x 0.0064/1.5 x 10^-3
C = 3.77 x 10^-11F
C = 37.77 pF ............Ans.

(b) we know that Q = CV

given that Q = 420x10^-6C and C from the above =  3.77 x 10^-11F

hence V = 420x10^-6C /  3.77 x 10^-11F = 11.14 x 10^6 V.........Ans.

(c)  Parallel plate cap
C = r(A/d) in Farads
is 8.8542e-12 F/m
r is dielectric constant = 22.8
A and d are area of plate in m² and separation in m

hence C = 3.77 x 10^-11F * 22.8 = 85.956 x 10^-11 F

Q is same as charge is conserved = 420uC

V = Q/C = 420x10^-6C / 85.956 x 10^-11F = 4.88 x 10^5 V.........Ans.

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