Respuesta :
Answer:
10000 V
0.00225988700565 m²
[tex]8\times 10^{-12}\ F[/tex]
Explanation:
E = Electric field = [tex]4\times 10^6\ V/m[/tex]
d = Gap = 2.5 mm
Q = Charge = 80 nC
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
Potential difference is given by
[tex]V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V[/tex]
The potential difference between the plates is 10000 V
Area is given by
[tex]A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2[/tex]
The area of the plate is 0.00225988700565 m²
Capacitance is given by
[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F[/tex]
The capacitance is [tex]8\times 10^{-12}\ F[/tex]
The plates of a parallel-plate capacitor which are placed in vacuum has the potential difference 10 kV, area [tex]2.26\times10^{-3}\rm m^2[/tex] and capacitance [tex]8\times10^{-12}\rm F[/tex].
What is capacitance of capacitor?
The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor can be given as,
[tex]C=\dfrac{Q}{V}[/tex]
Here, (Q) is the electric charge and (V) is the potential difference.
- (a) The potential difference between the plates-
The potential difference between the plates is the product of the magnitude of the electric field and the distance between the plates. Therefore,
[tex]V=4.00\times10^6\times0.0025\\V=10000\rm V\\V=10kV[/tex]
- (b) The area of each plate-
The magnitude of the electric field for the capacitor can be given as,
[tex]E=\dfrac{Q}{\varepsilon_o\times A}[/tex]
Here [tex]\varepsilon_o[/tex] is the permittivity which has the value ([tex]8.85\times10^{-12}[/tex]). As the charge of magnitude 80.0 nC. Thus, put the values in the above formula as,
[tex]4\times10^6=\dfrac{80\times10^{-9}}{8.85\times10^{-12}\times A}\\A=2.26\times10^{-3}\rm m^2[/tex]
- (c) The capacitance-
The capacitance of a capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. Thus,
[tex]C=\dfrac{80\times10^{-9}}{10000}\\C=8\times10^{-12}\rm F[/tex]
Hence, the plates of a parallel-plate capacitor which are placed in vacuum has,
- (a) The potential difference between the plates is 10 kV.
- (b) The area of each plate is [tex]2.26\times10^{-3}\rm m^2[/tex]
- (c) The capacitance is [tex]8\times10^{-12}\rm F[/tex]
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