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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top