Question

A parallel plate, air filled capacitor, has plates of area 0.82 m² and a separation of 0.040 mm.
(a) Find the capacitance.  
(b) If a voltage of 25 volts is applied to the capacitor, what is the magnitude of the charge on each plate?  
(c) What is the energy stored in the capacitor?  

Now the plates are filled with strontium titanate having a dielectric constant of 310 and a dielectric strength of 8.0 kV/mm.
(d) What is the new value of capacitance?  
(e) For a voltage of 25 V, what is the charge on each plate?  
(f) What is the energy stored in the capacitor now?  
(f) What is the maximum voltage that can be placed across the capacitor?

Answer 1

The capacitance of a parallel plate capacitor is given by

Where A is area of the plates, k is dielectric constant of the material between the plates ( k = 1 for air), d is separation between the plates and is permittivity of free space.

a.

b.

Total charge on this capacitor, when connected to 25 V, is

c.

The energy stored in this capacitor is

d.

Now Strontium titanate has been introduced between tha plates of the capacitor. It has a dielectric constant of 310.

So, the new capacitance of the capacitor is

e.

The charge on this new capacitor, when connected to 25 V, is

​​​​​​f.

Energy stored in this new capacitor is

g.

The dielectric strength of Strontium titanate is 8.0 kV /mm and the thickness of the dielectric layer is 0.040 mm. Therefore, it can withstand a maximum voltage of