Menu Close
Question 9: Derive an expression for the capacitance of a parallel plates capacitor when a dielectric is inserted between the plates of the capacitor.

Consider a parallel plate’s capacitor with comparatively large plates each of area A with a small separation d. In such a case electric field intensity is uniform between the plates and fringing are negligible. Let the medium between the plates is vacuum or air. Now suppose the capacitor is charged to an amount of charge Q; therefore, the surface charge density (charge per unit area) is
 Let the potential difference between the plates is V. Using Gauss Law
and put the value of σ, the electric intensity between the plates ise72c11p12 Now using the following equations;e73c11p12 Put the value of E from equation (1)e74c11p12 We know that capacitance of a capacitor is C = Q/V. Therefore, put the value of V (from the above) in this equation.e75c11p12 Capacitance when dielectric is inserted Let a dielectric of relative permittivity εr is inserted between the plates of the capacitor. Then the capacitance of the capacitor increases by a factor εr.e76c11p12 This equation shows that the capacitance depends upon the area of the plates A, their separation d and medium between them. Larger plates, closer they are and higher the dielectric constant of the medium between the plates, the greater will be the capacitance of the capacitor. Now let’s divide (iii) by (ii),e77c11p12 The relative permittivity is the ratio of the capacitance’s of a capacitor with a given material filling the space between the two plates to the capacitance of the same capacitor when the space is evacuated. This is also called dielectric constant or specific inductive capacity.


  1. Pingback:Electrostatics, Physics 12 … msas – msa

  2. Pingback:charging-discharging-of-capacitor – msa

  3. Pingback:electric-polarization-and-capacitance – msa

Leave a Reply

Your email address will not be published. Required fields are marked *