Respuesta :
Answer:
Explanation:
Suppose
Magnitude of Electric Field is E V/m
Area of the cross-section is A
capacitor [tex]C=\frac{\epsilon A}{d}[/tex]
Distance between Area of capacitor is d
Maximum Charge stored is
[tex]Q_{max}=capacitor\times Potential\ Difference[/tex]
[tex]Potential\ Difference=electric\ Field\times distance=E\times d[/tex]
[tex]Q_{max}=C\times Ed=CEd[/tex]
[tex]Q_{max}=\frac{\epsilon _0A}{d}\times Ed[/tex]
[tex]Q_{max}=\epsilon _0AE[/tex]
The potential difference is the ratio of charge and voltage. The potential difference across its plates will be [tex]\rm V= \frac{Qd}{\epsilon_0 A}[/tex].
What is a parallel plate capacitor?
It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have 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 capacitance for the parallel plate capacitor;
[tex]\rm C= \frac{\epsilon_0A}{d} \\\\[/tex]
The maximum value of the charge that the capacitor can store is given by;
[tex]\rm Q_{max}= C V \\\\ V = \frac{Q}{C} \\\\ \ V = \frac{Q}{\rm \frac{\epsilon_0A}{d} \\\\} \\\\[/tex]
[tex]\rm V= \frac{Qd}{\epsilon_0 A}[/tex]
Hence the potential difference across its plates will be [tex]\rm V= \frac{Qd}{\epsilon_0 A}[/tex].
To learn more about the parallel plate capacitor refer to the link;
https://brainly.com/question/12883102
