How Do Different Dielectrics Affect Capacitance in a Parallel-Plate Capacitor?

[peaceinmideas]

Homework Statement

There is a parallel-plate capacitor with a plate area A = 5.52 cm2 and a plate separation d = 5.52 mm. The left half of the gap is filled with material of dielectric constant κ1 = 7.00; the right half is filled with material of dielectric constant κ2 = 15.0. What is the capacitance?

Homework Equations

C= K(epsil) A/ d

The Attempt at a Solution

I tried to add the two values of K, I also converted both A and d to m^2 and m, respectively. I got 9.74 x 10^-14 F

 

[Redbelly98]

I'm getting a similar answer, except it disagrees in the 10^-14 part. I suspect the conversions to m^2 and m may be off; what did you get for A and d?

 

[tomcue]

as the capacitance. Is this correct?


Your attempt at a solution is on the right track. However, there is one mistake in your calculation. When adding the two values of K, you need to take into account the proportion of the gap that is filled with each material. In this case, since the left half of the gap is filled with material of dielectric constant κ1 = 7.00 and the right half is filled with material of dielectric constant κ2 = 15.0, the total capacitance would be calculated as follows:

C = (K1 * A/2 * d) + (K2 * A/2 * d)
= (7.00 * 5.52 cm^2/2 * 5.52 mm) + (15.0 * 5.52 cm^2/2 * 5.52 mm)
= (0.0385 m^2 * 0.00552 m) + (0.083 m^2 * 0.00552 m)
= 2.13 x 10^-7 m^3 + 4.58 x 10^-7 m^3
= 6.71 x 10^-7 m^3

Therefore, the capacitance of the parallel-plate capacitor is 6.71 x 10^-7 F. This is approximately 1.18 times larger than the value you calculated.

It is important to note that in this case, the dielectric constant values are not simply added, but rather the overall capacitance is calculated based on the proportion of the gap filled with each material. This is because the presence of different dielectric materials affects the electric field and thus the capacitance of the capacitor.