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](https://tex.z-dn.net/?f=A%20%3D%2010%20%5C%20cm%5E2)
![$=0.0010 \ m^2$](https://tex.z-dn.net/?f=%24%3D0.0010%20%5C%20m%5E2%24)
d = 10 mm
= 0.010 m
Then, Capacitance,
![$C=\frac{k \epsilon_0 A}{d}$](https://tex.z-dn.net/?f=%24C%3D%5Cfrac%7Bk%20%5Cepsilon_0%20A%7D%7Bd%7D%24)
![$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$](https://tex.z-dn.net/?f=%24C%3D%5Cfrac%7B8.85%20%5Ctimes%2010%5E%7B12%7D%20%5Ctimes%203%20%5Ctimes%200.0010%7D%7B0.010%7D%24)
![$C=2.655 \times 10^{12} \ F$](https://tex.z-dn.net/?f=%24C%3D2.655%20%5Ctimes%2010%5E%7B12%7D%20%5C%20F%24)
![$U_1 = \frac{1}{2}CV^2$](https://tex.z-dn.net/?f=%24U_1%20%3D%20%5Cfrac%7B1%7D%7B2%7DCV%5E2%24)
![$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$](https://tex.z-dn.net/?f=%24U_1%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%202.655%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%20%2815V%29%5E2%24)
![$U_1=2.987 \times 10^{-10}\ J$](https://tex.z-dn.net/?f=%24U_1%3D2.987%20%5Ctimes%2010%5E%7B-10%7D%5C%20J%24)
Now,
![$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$](https://tex.z-dn.net/?f=%24C_k%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bk%20%5Cepsilon_0%7D%7Bd%7D%20%5Ctimes%20%5Cfrac%7BA%7D%7B2%7D%24)
And
![$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$](https://tex.z-dn.net/?f=%24C_%7Bair%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5Cepsilon_0%7D%7Bd%7D%20%5Ctimes%20%5Cfrac%7BA%7D%7B2%7D%24)
In parallel combination,
![$C_{eq}= C_k + C_{air}$](https://tex.z-dn.net/?f=%24C_%7Beq%7D%3D%20C_k%20%2B%20C_%7Bair%7D%24)
![$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$](https://tex.z-dn.net/?f=%24C_%7Beq%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5Cepsilon_0%20A%7D%7Bd%7D%281%2Bk%29%24)
![$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$](https://tex.z-dn.net/?f=%24C_%7Beq%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B8.85%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%200.0010%7D%7B0.01%7D%20%5Ctimes%20%281%2B3%29%24)
![$C_{eq} = 1.77 \times 10^{-12}\ F$](https://tex.z-dn.net/?f=%24C_%7Beq%7D%20%3D%201.77%20%5Ctimes%2010%5E%7B-12%7D%5C%20F%24)
Then energy,
![$U_2 =\frac{1}{2} C_{eq} V^2$](https://tex.z-dn.net/?f=%24U_2%20%3D%5Cfrac%7B1%7D%7B2%7D%20C_%7Beq%7D%20V%5E2%24)
![$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$](https://tex.z-dn.net/?f=%24U_2%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%201.77%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%20%2815V%29%5E2%24)
![$U_2=1.99 \times 10^{-10} \ J$](https://tex.z-dn.net/?f=%24U_2%3D1.99%20%5Ctimes%2010%5E%7B-10%7D%20%5C%20J%24)
b). Now the charge on the
is :
![$Q=C_{eq} V$](https://tex.z-dn.net/?f=%24Q%3DC_%7Beq%7D%20V%24)
![$Q = 1.77 \times 10^{-12} \times 15 V$](https://tex.z-dn.net/?f=%24Q%20%3D%201.77%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%2015%20V%24)
![$Q = 26.55 \times 10^{-12} \ C$](https://tex.z-dn.net/?f=%24Q%20%3D%2026.55%20%5Ctimes%2010%5E%7B-12%7D%20%5C%20C%24)
Now when the capacitor gets disconnected from battery and the
is slowly
of the way out of the
is :
![$C_3=\frac{A \epsilon_0}{d}$](https://tex.z-dn.net/?f=%24C_3%3D%5Cfrac%7BA%20%5Cepsilon_0%7D%7Bd%7D%24)
![$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$](https://tex.z-dn.net/?f=%24C_3%20%3D%20%5Cfrac%7B0.0010%20%5Ctimes%208.85%20%5Ctimes%2010%5E%7B-12%7D%7D%7B0.01%7D%24)
![$C_3=0.885 \times 10^{-12} \ F$](https://tex.z-dn.net/?f=%24C_3%3D0.885%20%5Ctimes%2010%5E%7B-12%7D%20%5C%20F%24)
![$C_3 = 0.885 \times 10^{-12} \ F$](https://tex.z-dn.net/?f=%24C_3%20%3D%200.885%20%5Ctimes%2010%5E%7B-12%7D%20%5C%20F%24)
Without the dielectric,
![$U_3=\frac{1}{2} \frac{Q^2}{C}$](https://tex.z-dn.net/?f=%24U_3%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7BQ%5E2%7D%7BC%7D%24)
![$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$](https://tex.z-dn.net/?f=%24U_3%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B%2825.55%20%5Ctimes%2010%5E%7B-12%7D%29%5E2%7D%7B0.885%20%5Ctimes%2010%5E%7B-12%7D%7D%24)
![$U_3=3.98 \times 10^{-10} \ J$](https://tex.z-dn.net/?f=%24U_3%3D3.98%20%5Ctimes%2010%5E%7B-10%7D%20%5C%20J%24)
Answer:
Explanation:
In mirror generally, the type of image that can be formed by an object placed in front of it can either be a real image or a virtual image. Virtual images are images that are formed behind the mirror.
Virtual images are also images that cannot be formed from a photographic screen i.e the image formed by an object in front of the mirror cannot be reproduced compare to a real image than can be formed on a screen. This is the reason why we are able to get a picture of our self when pictures are taken using a photograph (For real images). This is not the same for virtual images.
We can only see ourselves when we stand in front of a mirror but in actual fact, we cannot reproduce the image of us we are seeing in the mirror.
<span>Any material that allows thermal energy to pass through easily is a conductor
</span>
Litmus paper and phenolphthalein
For the given portion of the diagram, the sequence will go as follows:Sedimentary rock ----> Heat and pressure ---->
metamorphic rock --->
melting ----> magma
Therefore, from the given choices, the correct choice would be:d. P represents metamorphic rocks and Q represents melting
Note: For the complete rock cycle, refer to the attached image.