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 separated by a 1.5mm air gap. What is the capacitance of tins 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 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

(c) Use C = Ke0A/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)

K =22.8
C = 22.8*8.85 x 10^-12 x 0.0064/1.5 x 10^-3
C = 0.8609 x 10^-9F
C = 860.928 pF

we know that Q = CV but it is not connected any voltage source hence Q =0 and V =0

(d) C is same = 860.928 pF

Q = CV if Q is 420x10^-6 C

V = 420x10^-6 / 860.928 pF = 4.87x10^5 Volt

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