Calculating Capacitance With Dielectric

[nishantve1]

Lets say There is a capacitor with the are of Plates A and the distance between the plates say is 5m if a dielectric of similar dimensions except the width is 3m is filled between the Capacitor plates how to calculate the new Capacitance ??

 

[DuckPaddle]

Imagine a very thin metal plate at the boundary of the two dielectrics. Now you have two capacitors, one with the dielectric and one without. Simply calculate the value of the two capacitors and use the capacitors in series formula to arrive at your answer.

 

[nishantve1]

Okay so let's say the capacitor plates are at a distance d from each other and the inserted dielectric is of x width and constant K the formula comes out to be KAE0 /Kd-x(K-1) which solves the problem thanks ! What is confusing me is if the dielectric is displaced towards one direction that is it does not touch any of the plates will it still be considered as two capacitors ?

 

[DuckPaddle]

Okay, so the dielectric is spaced off both plates. Just imagine a very thin metal plate on both sides of the dielectric, now we have three capacitors, calculate each one then calculate the total value for those three capacitors in series. Bottom line is when you encounter a boundary, just imagine a metal plate there splitting the capacitor in two. So if you have a problem where different dielectrics are stacked just divide the problem into so many individual capacitors, then figure them in the series circuit.

 

[sardar]

To calculate the new capacitance, we can use the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. In this scenario, the area of the plates (A) remains the same at 5m, but the distance between the plates (d) changes to 3m due to the insertion of the dielectric material.

To find the permittivity (ε) of the dielectric material, we can refer to a table or use the formula ε = ε0 * εr, where ε0 is the permittivity of free space (8.85 x 10^-12 F/m) and εr is the relative permittivity or dielectric constant of the material. Let's assume that the dielectric material has a relative permittivity of 2.5.

Plugging in the values, we get C = (8.85 x 10^-12 F/m * 2.5) * (5m/3m) = 14.75 x 10^-12 F = 14.75 pF (picofarads).

Therefore, the new capacitance of the capacitor with the dielectric material is 14.75 pF. This is an increase in capacitance compared to the original value without the dielectric material, which is expected as dielectric materials have a higher permittivity than air. The introduction of a dielectric material in a capacitor allows for a higher electric field to be sustained between the plates, resulting in an increase in capacitance.