#### Question

Two capacitors of capacitance 20⋅0 pF and 50⋅0 pF are connected in series with a 6⋅00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

#### Solution

Given :

`C_1 = 20.0 "pF"`

`C_2 = 50.0 "pF"`

When the capacitors are connected in series, their equivalent capacitance is given by `C_(eq) = (C_1C_2)/(C_1+C_2)`

∴ Equivalent capacitance,`C_(eq) = ((50 xx 10^-12) xx (20 xx 10^-12))/((50 xx 10^-12)+(20 xx 10^-12))` = `1.428 xx 10^-11 "F"`

(a) The charge on both capacitors is equal as they are connected in series. It is given by

`q = C_(eq) xx V`

⇒ `q = (1.428 xx 10^-11) xx 6.0 "C"`

Now ,

`V_1 = q/C_1 = ((1.428 xx 10^-11) xx 6.0 "C")/((20 xx 10^-12))`

⇒ `V_1 = 4.29 "V"`

and

`V_2 = (6.00 - 4.29) V = 1.71 "V"`

(b) The energies in the capacitors are given by

`E_1 = q^2/(2C_1)`

= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 20 xx 10^-12)`

= `184 "pJ"`

and

`E_2 = q^2/(2C_1)`

= `[(1.428 xx 10^-11) xx 6.0]^2` `xx 1/(2 xx 50 xx 10^-12)`

= `73.5 "pJ"`