Capacitors with dielectric question

[Daniiel]

[PLAIN]http://img340.imageshack.us/img340/4058/15244008.jpg

so
C = epysilon A / d
C= (8.85x10^-12 x 5.0x10^-3)/0.002 = 2.2125x10^-11
so the capacitance of the capacitors is ^
then with a dielectric where k = 3
can i use this equation

C = epysilon A k/ d, just putting the k next to epysilon, i just read that's what you do in my book
but yeh
when i do that i get C = 6.6375x10^-11
then i just added them as parralell capacitors
but the 6.6375x10^-11 doesn't seem right to me, that formula must be wrong, is it?
i thought the dielectric made the capacitance lower

 

[ehild]

No, the dielectric makes the capacitance higher. That is why a dielectric is applied!

ehild

 

[tiny-tim]

Hi Daniiel!

(have an epsilon: ε and try using the X² tag just above the Reply box )

… but the 6.6375x10^-11 doesn't seem right to me, that formula must be wrong, is it?
i thought the dielectric made the capacitance lower

No … the electric displacement field in the capacitor is the same, so the electric field (E = D/ε) is reduced, so the voltage is reduced (V = -∫Edx), so the capacitance (C = Q/V) is increased …

see eg http://en.wikipedia.org/wiki/Dielectric_constant" …

The dielectric constant is an essential piece of information when designing capacitors, and in other circumstances where a material might be expected to introduce capacitance into a circuit. If a material with a high dielectric constant is placed in an electric field, the magnitude of that field will be measurably reduced within the volume of the dielectric. This fact is commonly used to increase the capacitance of a particular capacitor design. The layers beneath etched conductors in printed circuit boards (PCBs) also act as dielectrics.​

 

[Daniiel]

ohh the voltageee is decreased
thanks guys
so that looks correct?

 

[rdl]

I would like to clarify that the formula you are using for capacitance with a dielectric is correct. The dielectric material between the plates of a capacitor will increase the capacitance by a factor of k, which is known as the dielectric constant. This is because the presence of the dielectric material reduces the electric field between the plates, allowing for more charge to be stored on the plates.

In your calculation, you correctly used the formula C = epysilon A k/ d to find the new capacitance with the dielectric material. However, it seems that you may have made a mistake in adding the two parallel capacitors together. When capacitors are connected in parallel, their total capacitance is the sum of their individual capacitances. Therefore, the total capacitance in this case would be C + C = 2C, not C + 6.6375x10^-11 as you have calculated.

In conclusion, the formula C = epysilon A k/ d is correct for calculating capacitance with a dielectric, and the presence of the dielectric does indeed increase the capacitance. I would suggest double-checking your calculations for the total capacitance when the capacitors are connected in parallel.