It is required to construct a 10μF capacitor which can be connected across a 200V battery. Capacitors of capacitance 10μF are available but they can withstand only 50V. Design a combination which can yield the desired result.

Hint: We have to design keeping in our mind the limitations that is the voltage. We are given a parallel plate capacitor. We know that the two plates are oppositely charged, the charges are equal in magnitude but opposite in polarity. and when connected to a source of emf, it conducts electricity but after it gets fully charged it stops

Complete step by step answer:Let the number of capacitors be m in one row.
Capacitance of each capacitor is 10 μF. So, total capacitance will be \[m\times 50=200\]
Solving this we get the value of m =4. So, the number of capacitors to be put in one single row are 4.
Now they are connected in series. We should find out the value of net capacitance in the series.
\[\begin{align}
  &\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}+\dfrac{1}{{{C}_{4}}} \\
 &\Rightarrow \dfrac{1}{C}=\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10} \\
 &\Rightarrow \dfrac{1}{C}=\dfrac{4}{10} \\
 &\Rightarrow C=\dfrac{10}{4} \\
\end{align}\]
Thus, the capacitance is \[\dfrac{10}{4}\] μF.
Now we assume there are t such rows. Thus,
\[\begin{align}
  & \dfrac{10}{4}\times t=10 \\
 &\Rightarrow t=4 \\
\end{align}\]
t=4 capacitors. So, the combinations of four rows each having 4 capacitors each will form a 10 μF capacitor that can withstand 200 V

Note:Apart from parallel plate capacitors, there exist a class of capacitors which are cylindrical in shape. Capacitors are added in series in the manner the resistances are added in parallel combination and vice versa. Also, we have not converted the capacitance in to Farads, as there was no such demand in the question.