Capacitors With and Without Dielectric

[jchoca]

Homework Statement

Two parallel-plate capacitors A and B are connected in parallel across a 620.0 V battery. Each plate has area 100 cm² and the plate separations are 4.5 mm. Capacitor A is filled with air; capacitor B is filled with a dielectric of dielectric constant 2.2.

A)Find the magnitude of the electric field within the dielectric of capacitor B.
B)Find the magnitude of the electric field within the air of capacitor A.
C)What is the free charge density on the higher-potential plate of capacitor A (with proper sign)?
D)What is the free charge density on the higher-potential plate of capacitor B (with proper sign)?
E)What is the induced charge density on the surface of the dielectric which is nearest to the higher-potential plate of capacitor B (with proper sign)?

Homework Equations

$$E=\frac{E_{0}}{E}\newline$$

$$E_{0}=\frac{\sigma}{\epsilon_{0}}\newline$$

$$V = Ed\newline$$

$$C_{eq}=C_{1}+C_{2}+\cdots+C_{n}$$

The Attempt at a Solution

The part that I am mainly having issues with (at the moment) is part A. I was able to obtain part B by using V = Ed and solving for E. Then I found part C by doing E₀=$$\sigma$$/$$\epsilon$$₀ and solving for $$\sigma$$. I tried to find part A by using K=E₀/E but that is incorrect. I'm sure it has something to do with the fact that there are two parallel capacitors rather than just one isolated one with a dielectric. I am pretty confused on how to approach getting part A.

 

[jchoca]

I made another attempt at the solution for A):

$$\frac{C}{C_{0}} = \frac{Q}{Q_{0}}=K$$

I know $$Q_{0}$$ since I know $$E_{0}$$.

$$Q=KQ_{0} =$$ 2.684 x $$10^{-8}$$C

$$\frac{Q}{A} = \epsilon_{0} =$$ 2.68 x $$10^{-6}$$ C/m²

$$E = \frac{\sigma}{\epsilon_{0}} =$$ 3.03 x $$10^{5}$$N/C

Still wrong though...

 

[baba89]

Firstly, it is important to understand the concept of capacitance and how it is affected by the presence of a dielectric material. Capacitance is a measure of the ability of a capacitor to store electric charge. It is directly proportional to the area of the plates and inversely proportional to the distance between them. When a dielectric material is inserted between the plates of a capacitor, it increases the capacitance by a factor of its dielectric constant, denoted by K.

Now, let's tackle part A. We know that the electric field within a capacitor is given by E = V/d, where V is the potential difference between the plates and d is the distance between them. In this case, the potential difference is 620 V and the distance is 4.5 mm (or 0.0045 m). Therefore, the electric field within the dielectric of capacitor B is given by:

E = 620 V/0.0045 m = 137,778 V/m

However, this is not the final answer as the dielectric material has a dielectric constant of 2.2. This means that the electric field within the dielectric will be reduced by a factor of 2.2. So, the final answer is:

E = 137,778 V/m / 2.2 = 62,626 V/m

Moving on to part B, we can use the same formula to find the electric field within the air-filled capacitor A. In this case, the electric field will be the same as the applied field of 620 V/0.0045 m = 137,778 V/m. This is because air has a dielectric constant of 1, so it does not affect the electric field.

For part C, we can use the formula E0 = σ/ε0 to find the free charge density on the higher-potential plate of capacitor A. Here, E0 is the electric field in air, which we found to be 137,778 V/m. The permittivity of free space, ε0, is a constant value of 8.85 x 10^-12 F/m. Therefore, we can rearrange the formula to find the free charge density, σ:

σ = E0 x ε0 = 137,778 V/m x 8.85 x 10^-12 F/m = 1.22 x 10^-7 C/m^2

This