Given,

Capacitances of the two capacitors, are 20.0 pF and 50.0 pF.

The voltage of the battery, V is 6V

Formula used

Let us represent the arrangement as

In a series arrangement the the charge on both the capacitance are same equal to total charge),can be found out by the equation,

Where Q and V represents the Charge and Potential difference respectively.

in series arrangement with Capacitance C₁ and C₂, C_eff can be found out as,

Or,

And thus the potential difference on each capacitance, V₁ and V₂ can be calculated by the below relations,

And,

Now,

The energy stored in a capacitor,E in Jules) can be found out by the relation,

Where

C is the capacitance of the capacitor in Farad

V is the potential difference across the capacitor.

a) First we calculate the ewuivalent capacitance by eqn.2.

Where C₁20 pF and C₂=50pF

So,

Hence, the total charge, Q from eqn.1 is

To find potential difference on each capacitor, we use eqn.3 and eqn.4. So the potential difference on 50pF capacitor is,

Or,

Similarly, on 20pF capacitor, V₂ is

Or,

Hence the potential differences across 50pF and 20pF capacitors are 1.714V and 4.29V respectively.

b) Energy stored in each capacitors can be calculat4ed by eqn.5

Hence for, 20pF capacitance across 4.29V potential difference, energy stored is,

Similarly for, 50pF capacitance across 1.71V potential difference, energy stored is,

Hence Energy stored in each capacitors are 73.5pJ and 184.04pJ for 50pF and 20pF capacitors respectively.