Physics, 27.07.2021 02:30 zanestone12

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. Express your answer numerically in joules.

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Answer from: westhacker1663

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 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 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. 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.

2. 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.

Solution :

Given :

$A = 10 \ cm^2$

$=0.0010 \ m^2$

d = 10 mm

= 0.010 m

Then, Capacitance,

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

$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$

$C=2.655 \times 10^{12} \ F$

$U_1 = \frac{1}{2}CV^2$

$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$

$U_1=2.987 \times 10^{-10}\ J$

Now,

$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$

And

$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$

In parallel combination,

$C_{eq}= C_k + C_{air}$

$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$

$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$

$C_{eq} = 1.77 \times 10^{-12}\ F$

Then energy,

$U_2 =\frac{1}{2} C_{eq} V^2$

$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$

$U_2=1.99 \times 10^{-10} \ J$

b). Now the charge on the $\text{capacitor}$ is :

$Q=C_{eq} V$

$Q = 1.77 \times 10^{-12} \times 15 V$

$Q = 26.55 \times 10^{-12} \ C$

Now when the capacitor gets disconnected from battery and the $\text{dielectric}$ is slowly $\text{removed the rest}$ of the way out of the $\text{capacitor}$ is :

$C_3=\frac{A \epsilon_0}{d}$

$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$

$C_3=0.885 \times 10^{-12} \ F$

$C_3 = 0.885 \times 10^{-12} \ F$

Without the dielectric,

$U_3=\frac{1}{2} \frac{Q^2}{C}$

$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$

$U_3=3.98 \times 10^{-10} \ J$

Answer from: Quest

in this image, you can see solar radiation going through ocean water.

no image, but probably absorption

Answer from: Quest

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

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