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. enter image description here

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

enter image description here

Which eqivalently is

enter image description here

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

enter image description here

Which can equivalently be considered to be

enter image description here

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

which is the correct answer

$$ $$ 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?


1 Answer 1


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 $d$.

In contrast, the fields in the top half of the structure in method 1 are very dependent on the physical presence of the lower half of the structure (and vice versa). So there you cannot analyze the two halves in isolation.

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).


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