A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 8.00cm on a side and 2.10mm thick. One slab is pyrex glass and the other is polystyrene1) If the potential difference between the plates is 76.0V , how much electrical energy is stored in the capacitor?U =_____

Respuesta :

Answer:

[tex]U=1.29\times 10^{-7}\ J[/tex]

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm  ,K₁= 4.7

d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

[tex]C_1=\dfrac{K_1\varepsilon _oA}{d_1}[/tex]

[tex]C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}[/tex]

[tex]C_1=1.26\times 10^{-10}\ F[/tex]

[tex]C_2=\dfrac{K_2\varepsilon _oA}{d_2}[/tex]

[tex]C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}[/tex]

[tex]C_2=0.701\times 10^{-10}\ F[/tex]

Net capacitance

[tex]C=\dfrac{C_1C_2}{C_1+C_2}[/tex]

[tex]C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F[/tex]

[tex]C=4.5\times 10^{-11}\ F[/tex]

We know that stored energy given as

[tex]U=\dfrac{CV^2}{2}[/tex]

V= 76 V

[tex]U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J[/tex]

[tex]U=1.29\times 10^{-7}\ J[/tex]

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