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rodikova [14]
3 years ago
15

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85x10-12 C2/N.m2 . Find the energy U1 of the dielectric-filled capacitor.
Physics
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

9.96\cdot 10^{-10}J

Explanation:

The capacitance of the parallel-plate capacitor is given by

C=\epsilon_0 k \frac{A}{d}

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2 is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F

Now we can calculate the energy of the capacitor, given by:

U=\frac{1}{2}CV^2

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

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