# Why do different ways of calculating the new capacitance of a capacitor after being partially filled with dialectric yield different results? [duplicate]

I have encountered this problem that stumped me and my colleagues. It is expected to be solved by Physics II students. I tried asking some faculty members but to no avail.

An air-filled parallel plate capacitor has capacitance of $$60$$ µF. Now one quarter of the space between the plates is filled with a dielectric constant $$\kappa = 5.0$$ as shown in the Figure. Find the new capacitance of the capacitor.

Using two different methods to divide the regions of the capacitor yield different values for the new capacitance.

Method 1

Let $$C_0$$ be the original capacitance of the capacitor

Divide the capacitor as such

Which eqivalently is

Then $$C_1 = \frac{\epsilon_0 A}{d/2} = 2 C_0$$

$$C_2 = \frac{\kappa \epsilon_0 A/2}{d/2} = \kappa C_0$$

$$C_3 = \frac{\epsilon_0 A/2}{d/2} = C_0$$

and $$C_{23} = C_2 + C_3 = C_0(\kappa +1)$$

$$\frac{1}{C_{eq}} = \frac{1}{C_{23}} + \frac{1}{C_1}$$

$$c_{eq} = \frac{C_1 C_{23}}{C_1 + C_{23}} = \frac{2(\kappa + 1)C_0}{\kappa +3} = 90$$ µF   Method 2

Divide capacitor as such

Which can equivalently be considered to be

Then $$C_1 = \frac{\epsilon_0 A/2}{d/2} = C_0$$

$$C_2 = \frac{\kappa \epsilon_0 A/2}{d/2} = \kappa C_0$$

$$C_3 = \frac{\epsilon_0 A/2}{d} = \frac{C_0}{2}$$

and $$\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\kappa C_0}{\kappa + 1}$$

Finally $$C_{eq} = C_{12} + C_3 = \frac{(3 \kappa + 1)C_0}{2(\kappa + 1)} = 80$$ µF

 I tried consulting Halliday's book (the assigned book for the course) but I did not find anything that cleared my confusion. Where have my colleagues and I make a mistake?

• In Method 1, essentially you have assumed that the horizontal line drawn from the top of the dielectric slab has the same voltage as the corresponding position in the empty right half. That is obviously too false to remedy. Commented Apr 23 at 9:55
• This exact question has been asked here many times. Please consider using the search function first. Commented Apr 23 at 10:01
• Commented Apr 23 at 10:03

Method 2 is the correct way to (approximately) analyze this situation, under the assumption that the width of the capacitor is much larger than the plate separation $$d$$ (which is also how your drawings depict it, so that seems to be OK).
The reason why method 2 is allowed is that the fields on the right-hand side are not very much changed by the physical presence of the left-hand side of the structure (and vice versa), if the width is large enough compared to $$d$$.
Of course if the width is not very large compared to $$d$$ you get more problems, even the simple plate capacitor formula would not hold any more due to the relative importance of fringing fields at the edges of the plates in that case. So even that would not help to make method 1 useful. (So for small width you really have to do 2D or 3D calculations).