Question

Two cylindrical capacitors of radii of inner and outer plates as(a,b) and (c,d)respectively. If they have same length and the same capacitance, then
$\begin{align}
  & a)\dfrac{a}{b}=\dfrac{c}{d} \\
 & b)a\times b=c\times d \\
 & c)\dfrac{a+b}{2}=\dfrac{c+d}{2} \\
 & d)a\times c=b\times d \\
\end{align}$

Answer 1

Hint: If we see the options provided, we can say that we are asked to obtain the relation between the inner and the outer radius of the above capacitors. We have to obtain the relation in such a way that the conditions mentioned in the question i.e. capacitance and the length of the capacitor is the same. Hence we will equate the capacitances of two such cylindrical capacitors and obtain the required relation.
Formula used:
$C=\dfrac{{{C}_{o}}}{L}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{r}{R} \right)}$

Complete answer:
Let us say there exists a cylindrical capacitor of length L with the radius of the inner cylinder as R and the radius of the outer cylinder as r. Let us say the region between the two cylinders consists of vacuum or air. Hence the capacitance of such capacitor per unit length is given by,
$C=\dfrac{{{C}_{o}}}{L}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{r}{R} \right)}$
Let us consider the first capacitor i.e. with inner radius a and outer radius b. Hence the capacitance ${{C}_{1}}$ of this capacitor is equal to,
$\begin{align}
  & {{C}_{1}}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{r}{R} \right)} \\
 & \Rightarrow {{C}_{1}}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{b}{a} \right)} \\
\end{align}$
Similarly, the capacitance ${{C}_{2}}$ of the capacitor with inner radius c and outer radius d. Hence the capacitance ${{C}_{2}}$ of this capacitor is equal to,
$\begin{align}
  & {{C}_{2}}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{r}{R} \right)} \\
 & \Rightarrow {{C}_{2}}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{d}{c} \right)} \\
\end{align}$
But since the capacitance of the two capacitors is equal we get,
$\begin{align}
  & {{C}_{2}}={{C}_{1}} \\
 & \Rightarrow \dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{d}{c} \right)}=\dfrac{2\pi {{\in }_{\circ }}}{{{\log }_{e}}\left( \dfrac{b}{a} \right)} \\
 & \Rightarrow {{\log }_{e}}\left( \dfrac{b}{a} \right)={{\log }_{e}}\left( \dfrac{d}{c} \right)\text{, }\because {{\log }_{e}}A={{\log }_{e}}B\Rightarrow A=B \\
 & \Rightarrow \dfrac{b}{a}=\dfrac{d}{c} \\
 & \Rightarrow \dfrac{a}{b}=\dfrac{c}{d} \\
\end{align}$

Hence the correct answer of the above question is option a.

Note:
It is to be noted that the above expression for capacitance is for capacitance per unit length of the capacitor. In the question it is already given to us that the length of the either capacitors is the same. Hence we could say that ${{C}_{2}}={{C}_{1}}$. It is also to be noted both the capacitors should also be placed in the same medium. Or the above condition obtained will not be valid.