Respuesta :
Answer:
U0/2
Explanation:
The potential energy stored by a capacitor is given by:
[tex]U=\frac{1}{2}CV^2[/tex] (1)
where
C is the capacitance
V is the potential difference applied across the capacitor
The capacitance of the parallel plate capacitor, at the beginning, is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where [tex]\epsilon_0[/tex] is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. Substituting this into eq.(1), we can write the initial potential energy stored in the capacitor as
[tex]U_0 = \frac{1}{2} \frac{\epsilon_0 A}{d}V^2[/tex]
Later, the separation between the plates is doubled:
[tex]d'=2d[/tex]
while the potential difference is kept constant. Therefore, we can calculate the new potential energy:
[tex]U' = \frac{\epsilon_0 A}{d'}V^2 = \frac{\epsilon_0 A}{2d}V^2 = \frac{U_0}{2}[/tex]
So, we see that the potential energy has been halved.
The energy stored in a capacitor is the electric potential energy, which is proportionate to the voltage of the capicitor and charge. Hence, as a result, the potential energy is half when the distance between the plate is half.
what is parrallel plate capicitor ?
It is a type of capacitor in which two metal plates are arranged in such a way that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates helps to stop the flow of electric current through it due to its non-conductive nature.
The potential energy held by a capacitor may be calculated as follows:
[tex]\rm{U = \frac{1}{2} Cv^{2} }[/tex]
The capacitance is denoted by the letter C.
The potential difference across the capacitor is denoted by V.
At first, the capacitance of the parallel plate capacitor is given by
[tex]C = \frac{\varepsilon_0A}{d}[/tex]
[tex]\rm{U = \frac{\varepsilon_0A}{d}v^{2} }[/tex]
According to the condition, the distance between the plates is later doubled d' = 2d
while keeping the potential difference constant As a result, we can compute the new potential energy as follows:
[tex]\rm{U' = \frac{\varepsilon_0A}{2d}v^{2} }[/tex]
[tex]U' = \frac{U}{2}[/tex]
Hence, as a result, the potential energy has been cut in half.
To learn more about the parallel plate capacitor refer to the link;
https://brainly.com/question/9324478
