Q. Two air capacitors $A = 1 \, \mu F; B = 4 \, \mu F$ are connected in series with $35 \, V$ source. When a medium of dielectric constant $K = 3$ is introduced in between the plates of $A$, charge on the capacitors changes by
Solution:
Initial situation is shown in the figure .
Net capacitance, $C = \frac{AB}{A + B} = \frac{4}{5} \mu F$
Charge through the circuit, $Q = CV$
$ = \frac{4}{5} \times 10^{-6} \times 35 = 28 \, \mu C$
When dielectric of dielectric constant $K = 3$ is introduced in between the plates of $A$
So its capacitance, $A'= KA = 3 \, \mu F$
Now, $A'$ and $B$ are in series, so net capacitance of the circuit,
$C' = \frac{A'B}{A' + B} = \frac{3 \times4}{3 + 4} = \frac{12}{7}\mu F$
Charge through the circuit
$ Q' = C'V = \frac{12}{7}\times 10^{-6} \times 35 = 60 \mu C$
Change in the charge on the capacitors
$= Q' - Q = 60 - 28 = 32\, \mu C. $
