Two capacitors each having a capacitor $C$ and breakdown voltage $V$ are joined in series. The effective capacitance and maximum working voltage of the combination is:
(A) $2C,2V$
(B) $\dfrac{C}{2},\dfrac{V}{2}$
(C) $2C,V$
(D) \[\dfrac{C}{2},2V\]

Hint: We know for a given capacitor, charge \[Q\] on a capacitor is proportional to potential difference \[V\], between the plates.
\[Q = CV\]
And, for series combination charges on capacitors remain the same.

Complete Step by Step Answer
Figure, shows two capacitors connected in series. The capacitance is \[C\] and \[C\].

Now, let us take the potential of the right plate of the second plate to be zero. The potential of the left plate of the first capacitor is \[E\]. Since, the breakdown voltage of capacitors is \[V\]. Therefore capacitor \[1\],
\[E - V = \dfrac{Q}{C}...........(i)\]
Similarly, for other capacitor,
\[V - 0 = \dfrac{Q}{C}...........(ii)\]
Adding equation \[(i)\] and \[(ii)\]
\[E = Q\left( {\dfrac{1}{C} + \dfrac{1}{C}} \right)\,.........\,(iii)\]
If the equivalent capacitance of the combination is \[{C_{eq}}\].
\[{C_{eq}} = \dfrac{Q}{E}\,..........\,(iv)\]
Using equation \[(iii)\] and \[(iv)\] we get,
\[{C_{eq}} = \dfrac{C}{2}\,\]
And, the maximum working voltage is \[E\].
Hence, \[E = V + V = 2V\]

Hence, Option (D) is correct

Note:
Charge on series combination remains same but voltage changes with respect to the capacitance whereas voltage on parallel combination remains same but charge varies in accordance to capacitance.