I've learnt that the capacitance $C$ of a parallel plate capacitor is given by: $$C=\frac{A\epsilon_0}{d}$$ where, $A$ is the area of cross section of the plates and $d$ is the separation between the two plates. When the space in between the two plates is filled by a dielectric of dielectric constant $K$ the new capacitance is given by: $$C'=KC=K\frac{A\epsilon_0}{d}$$ When the space between the plates is filled by a dielectric of dielectric constant $K$, the capacitance is increased by a factor of $K$. Is this applicable for all types of capacitors (spherical, cylindrical, etc.)? If yes, what is the reason behind this fact? For example, the capacitance of a spherical capacitor is given by:

$$C=\frac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1}$$

where, $r_1$ and $r_2$ are the radii of inner and outer metallic shells respectively. If we fill the entire region between the capacitor with a dielectric of dielectric constant $K$ will the resultant capacitance be given by:

$$C'=KC=K\frac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1}\ \ ?$$


1 Answer 1


. . . . the capacitance of a spherical capacitor is given by $C=\dfrac{4\pi\epsilon_0 r_1 r_2}{r_2-r_1} . . . .$

If $r_2 - r_1 = d$ and $d\ll r_1$ and $r_2$ then $4\pi r_1r_2$ is approximately equal to the surface area of a sphere $A$.

The capacitance of a spherical capacitor with these characteristics can now be written as $C=\dfrac{A\epsilon_0}{d}$ which is the capacitance of a parallel plate capacitor which can be thought of as a small part of a spherical capacitor.

Introducing a dielectric between the plates of a capacitor results in the dielectric being polarised ie the atoms being distorted so electric dipoles are set up.
These dipoles produce electric field which are in the opposite direction to the electric field which produced them and so the overall electric field is reduced which in turn results in a increase in the capacitance of a capacitor.
This effect does not depend on the type of capacitor which is being considered and so your factor $K$.

  • $\begingroup$ Thank you for your answer. In the first part, you've considered a special case where difference between the radii of the shells is much less than the two radii. I understood your method. But how can we tell the same fact is applicable for all conditions other than "the parallel limit"? $\endgroup$
    – Vishnu
    Jan 17, 2020 at 6:42
  • $\begingroup$ @GuruVishnu My purpose was to illustrate that two seeming different configurations of plates are essentially the same. Capacitance id defined as the ratio of the charge stored on the plates to the potential difference across the plates with no limitation as to the configuration of the plates. $\endgroup$
    – Farcher
    Jan 17, 2020 at 6:56
  • $\begingroup$ Fine. I thought the different orientations of the dipoles in the dielectric would alter the capacitance in a much different way. $\endgroup$
    – Vishnu
    Jan 17, 2020 at 7:00

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