Question

A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.

Answer 1

Answer:

A) The expression of the electric field halfway between the plates, if the plates are in the plane Y-Z is:

$\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}$

B) The expression for the magnitude of the electric field E₂ just in front of the plate two ends is:

$|E_2|=\displaystyle \frac{q}{2LW\varepsilon_0}$

C) The charge density is:

$\sigma_2=-4.2517\cdot10^{-3}C/m^2$

Completed question:

A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.

A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates

B) Input an expression for the magnitude of the electric field, E₂. Just in front of plate two END

C) If plate two has a total charge of q =-1 mC, what is its charge density, σ in C/m2?

Explanation:

A) The expression of the field can be calculated as the sum of the field produced by each plate. Each plate can be modeled as 2 parallel infinite metallic planes. Because this is a capacitor connected by both ends to a battery, the external planes have null charge (the field outside the device has to be null by definition of capacitor). This means than the charge of each plate has to be distributed in the internal faces. Because this es a metallic surface and there is no external field, we can consider a uniform charge distribution (σ=cte). Therefore in this case for each plane:

$\sigma_i=\displaystyle \frac{q_i}{LW}$

The field of an infinite uniform charged plane is:

$\vec{E_i}=\displaystyle \frac{\sigma_i}{2\varepsilon_0}sgn(x-x_{0i})\vec{x} =\frac{q_i}{2LW\varepsilon_0}sgn(x-x_{0i})\vec{x}$

In this case, inside the capacitor, if the plate 1 is in the left and the plate 2 is in the right, the field for 0<x<d is:

$\vec{E_1}\displaystyle=\frac{q}{2LW\varepsilon_0}sgn(x})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}$

$\vec{E_2}\displaystyle=\frac{-q}{2LW\varepsilon_0}sgn(x-d})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}$

$\vec{E}=\vec{E_1}+\vec{E_2}$

$\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}$

B) we already obtain the expression of the field E₂ inside the space between the plates. Even if we are asked the expression just in front of the plate and not inside, the expression for |E₂| is still de same.

C) As seen above, we already obtain the charge density expression. Therefore we only have to replace the variables for the numerical values.

Answer 2

Answer:

Explanation:

Check attachment