Two parallel plates, each having area A = 3490cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.36cm. You may assume (contrary to the drawing) that the separation between the plates is small compared with either linear dimension of the plate. A dielectric having dielectric constant κ = 3.6 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3490 cm2 and thickness equal to half of the separation (= 0.18 cm) . What is the charge on the top plate of this capacitor?

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SammiePreDuf SammiePreDuf

11-06-2016
Physics

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Two parallel plates, each having area A = 3490cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.36cm. You may assume (contrary to the drawing) that the separation between the plates is small compared with either linear dimension of the plate. A dielectric having dielectric constant κ = 3.6 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3490 cm2 and thickness equal to half of the separation (= 0.18 cm) . What is the charge on the top plate of this capacitor?

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akshayrao091 akshayrao091 · Answer 1 · 2016-06-22T12:18:42+02:00

the capacitance of a parallel plate capacitor partially filled with dielectric medium of dielectric constant K is =

C=ε₀A/(d-t+t/K)

A= area of each plate=34.9m
d= separation between plates=0.0036
t= thickness of dielectric medium =0.0018m
ε₀= electric constant (ε₀≈ 8.854×10^−12 F⋅m−1)
K= dielectric constant= 3.6

Q=CV
V is the potential difference or voltage of battery = 6v
Q is the charge over the plate
Therefore Q=8*10^-7 coulomb