How do I find the electric fields for this capacitor?

In summary, the formula for the electric field in a capacitor is given by E = V/d. By using this formula, we can obtain the electric field strengths in three different mediums, E1, E2, and E3, where d is equal in all three mediums. The total voltage can be calculated by adding the voltages of each medium. Next, we can find the boundary conditions between medium 1 and medium 2, which are both dielectric, and between medium 2 and medium 3, which is a dielectric and a conductor. The electric field strength values can be calculated using the equation E=Ps/e, derived from Gauss's Law, and then used to calculate the voltage and capacitance.
  • #1
Boltzman Oscillation
233
26
Homework Statement
The figure contains a capacitor consisting of three parallel conducting rectangular plates each of surface A, width W,. The plates are perfect dielectric conductors and non magnetic. The top and middle plates are separated by a distance 2d and filled with two dielectric layers. The middle and bottom plates are separated by a distance d and filled with air. The bottom plate i on the x-y plane. The voltage across the top and bottom plates is Vc. Determine the electric field intensity inside medium 1, 2, and 3. Determine the electric potential inside the capacitor.
Relevant Equations
Given below:
the image is given here along with some numerical information:
1571734309405.png


Now I know that the formula for the electric field in a capacitor is given as:
$$E = \frac{V}{d}$$
which I can use to obtain the three following fomulas:
$$E_1 = \frac{V_1}{d}$$
$$E_2 = \frac{V_2}{d}$$
$$E_3 = \frac{V_3}{d}$$
where I have used the fact that d is equal in all three mediums.
The total voltage is simply the addition of all of their voltages: V = V1 + V2 + V3.
Now I will attempt to find my boundary conditions.
Between medium 1 and medium 2 since they are both dielectric then the following is true:
$$\epsilon_1 E_{1n} - \epsilon_2 E_{2n} = p_s$$
Now I need the boundary conditions between medium 2 and medium 3.
Since there is a plate between medium 2 and 3 I will use the boundary conditions for a dielectric (medium 2) and conductor.
$$\epsilon_1 E_2 = \hat{n}p_s$$
I do not understand what to do now, am I working this out correctly or am I missing something. All help is appreciated.
 
Physics news on Phys.org
  • #2
I think an easy way to solve this problem is as follows : -

Assuming no fringe effect, the electric flux is uniformly distributed within the structure, then you can find the electric field strength values in those three mediums immediately by the equation E=Ps/e which can be derived from Gauss's Law.

E - electric field strength
Ps - surface charge density
e - permittivity

Once E1, E2 and E3 are obtained, all voltage and capacitance can be calculated in turn.
 

FAQ: How do I find the electric fields for this capacitor?

1. How do capacitors create an electric field?

Capacitors create an electric field by storing opposite charges on two parallel plates, which are separated by a dielectric material. The electric field is strongest between the plates and decreases as the distance from the plates increases.

2. What factors affect the strength of the electric field in a capacitor?

The strength of the electric field in a capacitor is affected by the distance between the plates, the magnitude of the charges on the plates, and the type of dielectric material used. The closer the plates are together and the larger the charges, the stronger the electric field will be. Different dielectric materials have different permittivity, which can also impact the strength of the electric field.

3. How do I calculate the electric field strength for a specific capacitor?

The electric field strength for a specific capacitor can be calculated using the formula E = V/d, where E is the electric field strength in volts per meter, V is the voltage across the plates in volts, and d is the distance between the plates in meters. This formula assumes that the plates are parallel and that there is a vacuum or air between them.

4. Can the electric field in a capacitor be changed?

Yes, the electric field in a capacitor can be changed by altering the distance between the plates, changing the magnitude of the charges on the plates, or using a different dielectric material. The direction of the electric field can also be changed by reversing the polarity of the charges on the plates.

5. How do I use the electric field to calculate the capacitance of a capacitor?

The electric field can be used to calculate the capacitance of a capacitor using the formula C = Q/V, where C is the capacitance in farads, Q is the charge on one of the plates in coulombs, and V is the voltage across the plates in volts. This formula assumes that the electric field is constant and uniform between the plates.

Similar threads

Replies
4
Views
1K
Replies
12
Views
1K
Replies
26
Views
2K
Replies
4
Views
2K
Back
Top