Two parallel plates, each having area A = 2870cm 2 are connected to the terminal

Question

Two parallel plates, each having area A = 2870cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.48cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

1) What is C, the capacitance of this parallel plate capacitor?

2) What is Q, the charge stored on the top plate of the this capacitor?.

3) A dielectric having dielectric constant = 3.1 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 2870 cm2 and thickness equal to half of the separation (= 0.24 cm) . What is the charge on the top plate of this capacitor?

Explanation / Answer

1) C = e0A/d = 8.85*10^-12*0.287/0.0048 = 529.16 pF

2)    Q = CV = 529.16*10^-12*6 = 3.17 nC

3) Capacitance for unfilled capacitor = C' = e0A/(d/2) = 2C = 1058.32 pF

Capacitance for filled capacitor = C'' = ke0A/(d/2) = kC' = 3280.792 pF

equivalent capacitance = Ceq = 800.2 pF

Now Charge = Q' = Ceq*V =800.2*10^-12 * 6 = 4.8 nC

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