Respuesta :
Answer:
C). [tex]U_f = \frac{U_0}{2}[/tex]
Explanation:
As we know that capacitance of a given capacitor is
[tex]C = \frac{\epsilon_0 A}{d}[/tex]
now we know that energy stored in the capacitor plates
[tex]U_0 = \frac{Q^2}{2C}[/tex]
here if all the dimensions of the capacitor plate is doubled
then in that case
[tex]C' = \frac{\epsilon_0 (4A)}{2d}[/tex]
here area becomes 4 times on doubling the radius and the distance between the plates also doubles
So new capacitance is now
[tex]C' = 2C[/tex]
so capacitance is doubled
now the final energy stored between the plates of capacitor is given as
[tex]U_f = \frac{Q^2}{2C'}[/tex]
so the final energy is
[tex]U_f = \frac{Q^2}{4C}[/tex]
[tex]U_f = \frac{U_0}{2}[/tex]
Option C is correct.ll the geometric parameters of the capacitor are now doubled.The original energy stored in the capacitor was U₀. Energy does it now store [tex]\rm \frac{U_0}{2}[/tex].
What is parallel plate capacitor ?
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 capacitance between the two plates is given by,
The capacitance of a given capacitor is;
[tex]\rm C=\frac{\epsilon_0A}{d}[/tex]
The energy stored in the capacitor plates is;
[tex]\rm U = \frac{Q^2}{2C}[/tex]
For the given new conditions;
[tex]\rm C'=\frac{\epsilon_0( 4A)}{2d}\\\\ \rm \rm C'=2C[/tex]
Energy stored is the inversely proportional to the capacitance then . If the original energy stored in the capacitor was U₀. Energy does it now store [tex]\rm \frac{U_0}{2}[/tex].
Hence option C is correct
To learn about the parallel-plate capacitor refer to the link;
https://brainly.com/question/20716849
