Respuesta :
1.8 × 10-6 C You need to know that charge Q=C*V, that the capacitance C=epsilon_0*A/d and that the electric field between plates is given by E=V/d. Now use these three equations to solve for Q, divide by 2 and get answer
The capacitor has the ability to store an electrical charge Q (units in Coulombs) of electrons. So the magnitude of the charge between the plates is [tex]\rm Q= 3.54\;\times\;10^{-6}\;\rm C[/tex].
Here is the explaination the given below.
A parallel-plate capacitor has a plate area of [tex]\rm A = 0.2\; \rm m^2\\[/tex] .
The separation between the plates [tex]\rm d = 0.1\;\rm mm[/tex].
The electric field between the plates is [tex]\rm E = 2.0\;\times\;10^6\;\rm {v/m}[/tex].
The magnitude of the charge on plates can be calculated as per the formula.
[tex]\rm Q= CV[/tex]
Where,
[tex]\rm C[/tex] is the capacitance between the plates.
[tex]\rm V[/tex] is the voltage across the plates.
The formula to calculate the capacitance is given below.
[tex]\rm C = \varepsilon \dfrac {A} {d}[/tex]
Where, Permittivity [tex]\varepsilon = 8.85\;\times\;10^{(-12)}[/tex]
By substituting the values in the formula to calculate the capacitance,
[tex]\rm C = 8.85\;\times\;10^{(-12)}\;\times(\dfrac {0.2}{0.1\;\times 10^{-3}})[/tex]
[tex]\rm C = 1.77\;\times\;10^{-8}[/tex]
The formula to calculate the voltage is given below.
[tex]\rm V = \rm Ed[/tex]
By substituting the values in the formula to calculate the voltage,
[tex]\rm V= 2.0\;\times\;10^6\;\times\;0.1\;\times\;10^{-3}[/tex]
[tex]\rm V= 200 \rm V[/tex]
So the charge will be calculated as,
[tex]\rm Q= CV[/tex]
[tex]\rm Q = 1.77 \times 10^{-8}\;\times\;200[/tex]
[tex]\rm Q= 3.54\;\times\;10^{-6}\;\rm C[/tex]
The magnitude of charge between the plates of the capacitor is [tex]3.54\;\times\;10^{-6}\;\rm C[/tex].
For more details about the capacitor, follow the link given below.
https://brainly.com/question/14048432.
