Question

A parallel plate air capacitor with no dielectric between the plates is connected to a constant voltage source. How would the capacitance and the charge change if a dielectric of dielectric constant K=2 is inserted between the plates. C0 and Q0 are the capacitance and charge of the capacitor before the introduction of the dielectric.

Answer 1

Here as we have given that,

Initially there was air in between the plates of the capacitor so that form the relation we know that the Value of capacitor is entirely depending on its geometry so that,

C0 = Ko A/d

Where ,

Ko= permittivity of the free space

A = area of the plates

d = distance between the plates

Now we are told that we insert a dielectric of dielectric constant k = 2 is inserted so that,

C' = Ko ×K ×A/d = 2 × Co

Hence the new dielectric will becomes 2 times the previous value of the capacitance Co

Hence C(new) = 2 Co

Now for the charge we know that,

We have a formula as,

Qo = CoV

Where V is the constant voltage source given

Now here when we insert a dielectric then here there is increase in the value of capacitor but the voltage remains same so that,

Q(new ) = 2Co × V = 2Qo

Which means now the value of the capacitor to store charges becomes twice.  

Or , Q(new) = 2Qo