What is the Capacitance of a Capacitor with Two Different Dielectrics?

Are the plates parallel to one another or perpendicular?In summary, the problem involves calculating the capacitance of a capacitor made of two square metal plates with dielectric constants of 4.61 and 6.9 in the upper and lower halves, respectively. The capacitance is found to be 2.935057735*10^-9 pF by using the formula C = C_top + C_bottom and plugging in the appropriate values. However, more information is needed about the plate size, separation, and orientation for a definitive answer.
  • #1
DrunkApple
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0

Homework Statement


A capacitor is constructed from two square
metal plates. A dielectric [itex]\kappa[/itex] = 4.61 fills the
upper half of the capacitor and a dielectric
[itex]\kappa[/itex] = 6.9 fills the lower half of the capacitor.
Neglect edge effects. Calculate the capacitance C of the device.
Answer in units of pF


Homework Equations


C[itex]_{TOP}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
C[itex]_{bottom}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
C[itex]_{total}[/itex] = C[itex]_{TOP}[/itex] + C[itex]_{BOTTOM}[/itex]

The Attempt at a Solution


C[itex]_{TOP}[/itex] = [itex]\kappa[/itex][itex]\epsilon[/itex]A/d
= 4.61(8.85419*10[itex]^{-12}[/itex])(.12[itex]^{2}[/itex])/0.0005
= 1.175553098*10[itex]^{-9}[/itex]
C[itex]_{bottom}[/itex] = 6.9(8.85419*10[itex]^{-12}[/itex])(.12[itex]^{2}[/itex])/0.0005
= 1.759504637*10[itex]^{-9}[/itex]
C[itex]_{total}[/itex] = 2.935057735*10[itex]^{-9}[/itex]
Is this correct?
 
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  • #2
DrunkApple said:

Homework Statement


A capacitor is constructed from two square
metal plates. A dielectric [itex]\kappa[/itex] = 4.61 fills the
upper half of the capacitor and a dielectric
[itex]\kappa[/itex] = 6.9 fills the lower half of the capacitor.
Neglect edge effects. Calculate the capacitance C of the device.
Answer in units of pF
Ummm. Plate size? Plate separation? Are the plates oriented vertically or horizontally?
 

What is capacitance?

Capacitance is the ability of a capacitor to store electric charge. It is measured in farads (F) and is represented by the letter C.

How is capacitance calculated?

The capacitance of a capacitor can be calculated by dividing the electric charge (Q) by the potential difference (V) between the plates. Therefore, C = Q/V.

What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by three main factors: the distance between the plates, the area of the plates, and the material between the plates. The closer the plates are, the larger the area of the plates, and the higher the permittivity of the material between the plates, the higher the capacitance will be.

What is the relationship between capacitance and voltage?

Capacitance and voltage have an inverse relationship. This means that as voltage increases, capacitance decreases, and vice versa. This is because as the potential difference between the plates increases, the electric field strength between them also increases, resulting in a decrease in capacitance.

How does a capacitor store energy?

A capacitor stores energy by storing electric charge on its plates. When a voltage is applied, electrons from one plate are attracted to the other plate, creating a potential difference between the plates. This potential difference represents stored energy in the capacitor. When the capacitor is discharged, the stored energy is released.

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