Respuesta :
To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.
Energy in a capacitor can be defined as
[tex]E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}[/tex]
Here,
V = Potential difference across the capacitor plates
Q = Charge stored on the capacitor plates
At the same time capacitance can be defined as,
[tex]C = \epsilon_0 (\frac{A}{d})[/tex]
Here,
[tex]\epsilon_0 =[/tex] Vacuum permittivity constant
A = Area
d = Distance
Replacing with our values we have that,
[tex]C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})[/tex]
[tex]C = 1.5901*10^{-10}F[/tex]
PART A) Energy stored in the capacitor is
[tex]E = \frac{1}{2} CV^2[/tex]
[tex]E = \frac{1}{2} (1.5901*10^{-10})(575)^2[/tex]
[tex]E = 2.628*10^{-5}J[/tex]
PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore
[tex]Q = CV[/tex]
[tex]Q = (1.59*10^{-10})(575)[/tex]
[tex]Q = 9.1425*10^{-8}C[/tex]
Now if [tex]d = 10.04*10^{-3}m[/tex] we have that the capacitance is
[tex]C = \epsilon_0 (\frac{A}{d})[/tex]
[tex]C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})[/tex]
[tex]C = 3.9754*10^{-11}F[/tex]
Then the energy stored is
[tex]E = \frac{1}{2} \frac{Q^2}{C}[/tex]
[tex]E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})[/tex]
[tex]E = 1.051*10^{-4} J[/tex]
PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore
[tex]W = 1.051*10^{-4} J -2.628*10^{-5}J[/tex]
[tex]W = 7.882*10^{-5} J[/tex]
Part A: The energy stored in the capacitor is [tex]2.628\times 10^{-5}\;\rm J[/tex].
Part B: The separation in the plates of the capacitor is 10.04 mm, in the case the energy stored is [tex]6.57\times 10^{-6}\;\rm J[/tex].
Part C: The total work done to increase the separation between the plates of the capacitor is [tex]1.97 \times 10^{-5}\;\rm J[/tex].
Energy of Capacitor
Given that the area of the parallel plate capacitor is 451 cm2 and air-filled gap between the plates is 2.51 mm. The potential difference across the capacitor is 575 V.
Part A
The energy stored in the capacitor is calculated by the equation given below.
[tex]E = \dfrac {1}{2} CV^2[/tex]
Here, E is the energy stored in the capacitor, C is the capacitance and V is the potential difference across the capacitor.
The capacitance of the capacitor is,
[tex]C = \varepsilon _\circ \dfrac{A}{d}[/tex]
Where A is the area and d is the distance between the plates of the capacitor. [tex]\varepsilon _\circ[/tex] is the permittivity constant that is [tex]8.85\times10^-12[/tex].
[tex]C = 8.85\times 10^{-12}\times \dfrac {451\times 10^{-4}}{2.51\times 10^{-3}}[/tex]
[tex]C = 1.59\times 10^{-10}\;\rm F[/tex]
Now, the energy stored in the capacitor is,
[tex]E = \dfrac {1}{2}\times 1.59\times 10^{-10}\times (575)^2 [/tex]
[tex]E = 2.628\times 10^{-5}\;\rm J[/tex]
Hence we can conclude that the energy stored in the capacitor is [tex]2.628\times 10^{-5}\;\rm J[/tex].
Part B
The separation between the plates is 10.04 mm. The capacitance will be given as,
[tex]C = \varepsilon _\circ \dfrac{A}{d}[/tex]
[tex]C = 8.85\times 10^{-12}\times \dfrac {451\times 10^{-4}}{10.04\times 10^{-3}}[/tex]
[tex]C = 3.975\times 10^{-11}\;\rm F[/tex]
The energy stored in this case is,
[tex]E = \dfrac {1}{2}\times 3.975\times 10^{-11}\times (575)^2[/tex]
[tex]E = 6.57\times 10^{-6}\;\rm J[/tex]
The separation in the plates of the capacitor is 10.04 mm, in the case the energy stored is [tex]6.57\times 10^{-6}\;\rm J[/tex].
Part C
The amount of work required to increase the separation between the plates of the capacitor is
Work = Difference between the energies stored in the capacitor
[tex]W = 2.628 \times 10^{-5} - 6.57 \times 10^{-6} \;\rm J[/tex]
[tex]W = 1.97\times 10^{-5}\;\rm J[/tex]
Hence we can conclude that the total work done to increase the separation between the plates of the capacitor is [tex]1.97 \times 10^{-5}\;\rm J[/tex].
To know more about the energy stored in the capacitor, follow the link given below.
https://brainly.com/question/14739936.
