Respuesta :
Answer:
D) Q/2
Explanation:
The amount of charge on a capacitor is given by:
Q = CV
where
C is the capacitance
V is the voltage
The capacitance of a parallel-plate capacitor is given by
[tex]C=\epsilon_0 \frac{A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
Substituting the last formula into the first one, we can write
[tex]Q=\epsilon_0 \frac{AV}{d}[/tex]
In this problem, the two plates are pulled apart to twice their original separation, so
d' = 2d
While the voltage V is kept constant. Therefore, the new charge stored in the capacitor will be
[tex]Q'=\epsilon_0 \frac{AV}{2d}=\frac{1}{2} \epsilon_0 \frac{AV}{d}=\frac{Q}{2}[/tex]
Option D is right.The amount of charge on the plates is now equal to Q/2.Charges are stores in the plate placed parallel referred as the parallel plate capacitor.
What is parallel plate capacitor ?
It is an type capacitor is an in which two metal plates arranged in such away so that they are connected in parallel and having some distance between them.
A dielectric medium is must in between these plates help to stop the flow of electric current through it due to its non-conductive nature .
The value of charge force the capacitor is given as:
Q = CV
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The value of capacitance of a parallel-plate capacitor is given by the formula;
[tex]\rm C= \epsilon_0\frac{AV}{d}[/tex]
If all the values are constant then the charge is inversely proportional the distance between the charge.
Hence if the distance is doubled in that condition the value of charge will also half.
d' = 2d
While the voltage V is kept constant. Therefore, the new charge stored in the capacitor will be
[tex]\rm Q'= \frac{Q}{2}[/tex]
Hence option D is right.The amount of charge on the plates is now equal to Q/2.
To learn more about the parallel plate capacitor refer to the link;
https://brainly.com/question/12883102
