How Does Capacitance Change with Multiple Dielectric Materials in a Capacitor

AI Thread Summary
Capacitance in capacitors with multiple dielectric materials can be calculated using a modified version of Maxwell's equations. The equivalent capacitance for capacitors with two distinct dielectrics, each with different dielectric constants, can be determined using the formula C_eq = (C_1 C_2) / (C_1 + C_2), where C_1 and C_2 are the capacitances of each dielectric. If the thickness of each dielectric is equal, the total distance between the plates is effectively doubled. Understanding this approach allows for the integration of multiple dielectric materials into the capacitance calculation. This method highlights the importance of dielectric constants and their arrangement within the capacitor.
xcutexboax
Messages
14
Reaction score
0
Hey Guys,

I was just doing a question on capacitance and i was wondering since capacitance is usually determined by a fixed formula which is dependent on the di electric material that is contained within a capacitor.. However it struck me that a capacitor does not neccesarily contain only one kind of material. How does the expression of the formula changes if a capacitor can contain more than one kind of di-electric materials... pls enlighten me... IS it based on the area they occupy? ThAnks. :confused:
 
Physics news on Phys.org
IIRC, for multiple dielectric capacitors, you would simply use the modification of one of Maxwell's equations, namely: V = \int{\kappa\overrightarrow{E} \cdot d\overrightarrow{s}. In this case, you integrate over the thickness of one capacitor, then over the thickness of the other capacitor, then add the two results to find the electric potential across the plates.
 
If I was referring to the formula C=k*epsilon*A/d where k is the dielectric constant of the material, how does finding the potential change the above expression? I mean if a capacitor can contain like 3 dielectric materials of different k, how does it affect the above expression/formula? =)
 
If you have a capacitor with two distinct materials as dielectrics in between the plates with dielectric constants \kappa_1 & \kappa_2 (and assuming you're dealing with a standard, flat parallel plate capacitor here), if the thickness of material one is the same as that of material two (let's say a thickness of d), then the equivalent capacitance is given by C_{eq} = \frac{C_1 C_2}{C_1+C_2}, where C_1 = \frac{\kappa_1 \epsilon_0 A}{d}, and same for C_2. Of course, in this case, I'm taking the actual plate-distance to be 2d. :wink:
 
Oh i finally understand... u took the capacitor like a circuit which contatins other "capacitors". Marvellous... Okie thanks for the tip.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top