A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2, plate separation d = 5.00mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 10.0V . Throughout the problem, use Eo = 8.85×10−12 C2/nxm2.

a) find the energy U1 of the dielectric filled capacitor

b)The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

c) The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

d)In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?

all answer need to be expressed numerically in joules,

b) When the dielectric is pulled out halfway with the battery still connected to the capacitor, then it acts like 2 capacitors connected in parallel.

Let the capacitance of the half with no dielectric be [tex]C_1[/tex] and of the filled half be [tex]C_2[/tex]. Then,

[tex]C_1=\frac{8.85\times10^{-12}\times15\times100^{-2}}{5\times10^{-3}}= 2.655\times10^{-12}F[/tex]

[tex]C_2=\frac{4\times8.85\times10^{-12}\times15\times100^{-2}}{5\times10^{-3}}= 1.062\times10^{-11}F[/tex]

Therefore, total capacitance [tex]C=C_1+C_2=1.328\times10^{-11}F[/tex]

and, [tex]U_2=\frac{1}{2} \times1.328\times10^{-11}\times10^2J=6.638\times10^{-10}J[/tex]

c) Let us first calculate the charge stored in the previous configuration,

[tex]Q=CV=1.328\times10^{-11}\times10C=1.328\times10^{-10}C[/tex]

Now, the dielectric is completely removed. Then the new capacitance is,

[tex]C=\frac{8.85\times10^{-12}\times30\times100^{-2}}{5\times10^{-3}}F=5.310\times10^{-12}F[/tex]

Therefore, energy of the capacitor, [tex]U_3=\frac{1}{2}\frac{Q^2}{C} =\frac{1}{2}\frac{(1.328\times10^{-10})^2}{5.310\times10^{-12}}J=1.661\times10^{-9}J[/tex]

d) The work done to remove the dielectric will be the difference in the energies of the 2 configurations (half-filled and fully removed), that is,

[tex]W=U_3-U_2=9.968\times10^{-10}J[/tex]

codiepienagoya codiepienagoya · Answer 2 · 2022-02-01T18:47:37+01:00

Following are solutions to the given points:

Given:

Please find the question.

To find:

find points=?

Solution:

For part A)

[tex]\to U_1 = 0.5\times k\times C \times V^2 \\\\\to C = e_o\times \frac{A}{d} = \frac{(8.85\times 10^{-12} \times 30 \times 10^{-4})}{(5\times 10^{-3})} = 5.31\times 10^{-12}\ F\\\\\to U_1 = 0.5\times 4\times 5.31 \times 10^{-12}\times 10^{2} =1.062\times 10^{-9}\ J\\\\[/tex]

For Part B)

when the capacitor is half-filled

[tex]\to C' = C_1+C_2 = k\times e_o\times \frac{A}{(2\times d)} = e_o\times \frac{A}{(2\times d)} = (e_o\times \frac{A}{d})\times (\frac{(k+1)}{2}) = \frac{(k+1) \times C}{2}\\\\\to U_2 = 0.5\times C'\times V^2 = 0.5\times ( \frac{(k+1)\times C}{2})\times V^2 \\\\ =0.5\times ( \frac{( (4+1)\times 5.31\times 10^{-12})}{2})\times 10^2\to U_2 = 6.6375 \times 10^{-10}\ J\\\\[/tex]

For Part C)

when disconnected from the battery the charge on the capacitor remains the same

[tex]\to U_3 = \frac{U_2}{k} = \frac{(6.6375\times 10^{-10})}{4} = 1.6593\times 10^{-10}\ J\\\\[/tex]

For Part D)

[tex]\to W = U_2-U_3 = (6.6375-1.6593)\times 10^{-10} = 4.9782\times 10^{-10}\ J[/tex]

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