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Hint: We know that $C=\dfrac{Q}{V}$, so find an expression of V in terms of Q (Q is the charge on the inner cylinder). Consider a Gaussian cylindrical surface of length y and use Gauss’s law that is $\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}$. Here ${{Q}_{enclosed}}=\dfrac{Qy}{L}$. Then use the formula $\phi =\int{\overrightarrow{E}.d\overrightarrow{s}}$ find the electric field in terms of Q, L. Finally, use ${{V}_{A}}-{{V}_{B}}=-\int\limits_{{{r}_{B}}}^{{{r}_{A}}}{\overrightarrow{E}.d\overrightarrow{r}}$ to find and expression for V in terms of Q, L and substitute this expression in $C=\dfrac{Q}{V}$

Formula used:

$C=\dfrac{Q}{V}$

$\phi =\int{\overrightarrow{E}.d\overrightarrow{s}}$

$\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}$

${{V}_{A}}-{{V}_{B}}=-\int\limits_{{{r}_{B}}}^{{{r}_{A}}}{\overrightarrow{E}.d\overrightarrow{r}}$

Capacitor is a device that stores electric energy in the form of an electric field by storing charge on its surface.

Capacitance C of a capacitor is defined as the magnitude of the charge Q on the positive plate by the magnitude of the potential difference V between the plates. Therefore, $C=\dfrac{Q}{V}$.

A cylindrical capacitor of length L consists of two cylinders of radii ${{R}_{1}}$ and ${{R}_{2}}$. Let ${{R}_{2}}$> ${{R}_{1}}$.

The outer cylinder is earthed. The cylinder is long enough so that we can neglect the fringing of the electric field at the ends. Electric field at a point between the cylinders will be radial and its magnitude will depend on the distance from the central axis.

Consider a Gaussian surface of length y and radius r such that ${{R}_{2}}$< r < ${{R}_{1}}$.

Flux through the plane surface is zero because electric field and area vectors are perpendicular to each other.

For curved part flux is equal to $\phi =\int{\overrightarrow{E}.d\overrightarrow{s}}=\int{Eds\cos \theta }$, where $\theta $ is the angle between vectors $\overrightarrow{E}$ and $d\overrightarrow{s}$.

Since E and ds are in the same direction, $\theta =0\Rightarrow \cos \theta =\cos 0=1$.

Therefore, $\phi =\int{Eds}$.

Since, E is constant (due to symmetry) on this surface, $\phi =E\int{ds}=EA$ …..(i).

A is the total curved surface area of the cylinder.

We know, $A=2\pi ry$. Substitute the value of A in equation (i).

Therefore, $\phi =EA=E.2\pi ry$ …….(ii).

Let the total charge on the surface of the inner cylinder be Q.

Charge inside the surface is equal to $q=\dfrac{Qy}{L}$.

From Gauss's law, $\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}$ we get,

$\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}=\dfrac{q}{{{\varepsilon }_{\circ }}}=\dfrac{\dfrac{Qy}{L}}{{{\varepsilon }_{\circ }}}=\dfrac{Qy}{L{{\varepsilon }_{\circ }}}$ ……(iii).

From equations (ii) and (iii) we get,

$\dfrac{Qy}{L{{\varepsilon }_{\circ }}}=E.2\pi ry$.

Therefore, $E=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}rL}$ ………(iv).

Potential difference between two points A and B is given by ${{V}_{A}}-{{V}_{B}}=-\int\limits_{{{r}_{B}}}^{{{r}_{A}}}{\overrightarrow{E}.d\overrightarrow{r}}$ .

Here, ${{V}_{A}}$ and ${{V}_{B}}$ are potential of the inner and outer cylinders respectively.

${{r}_{A}}={{R}_{1}}$ and ${{r}_{B}}={{R}_{2}}$.

Therefore,

${{V}_{A}}-{{V}_{B}}=V=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{\overrightarrow{E}.d\overrightarrow{r}}=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{Edr}$. (because E and dr are in the same direction).

Substitute the value of E from equation (iv) in the above equation.

Therefore,

$V=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}rl}.dr}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{\dfrac{dr}{r}}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\left. \ln r \right|_{{{R}_{1}}}^{{{R}_{2}}}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}$.

As discussed above, capacitance $C=\dfrac{Q}{V}=\dfrac{Q}{\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}}=\dfrac{2\pi {{\varepsilon }_{\circ }}L}{\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}}$.

Therefore, capacitance of a cylindrical capacitor is directly proportional to its length.

Hence the correct option is A.

Note: In the case of a cylindrical capacitor, charge is stored only on the inner cylinder. The outer cylinder is earthed so it will not have any charge on it. When a plate or conductor is being earth, the potential at that point or surface is considered as zero.

Formula used:

$C=\dfrac{Q}{V}$

$\phi =\int{\overrightarrow{E}.d\overrightarrow{s}}$

$\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}$

${{V}_{A}}-{{V}_{B}}=-\int\limits_{{{r}_{B}}}^{{{r}_{A}}}{\overrightarrow{E}.d\overrightarrow{r}}$

**Complete step-by-step answer:**Capacitor is a device that stores electric energy in the form of an electric field by storing charge on its surface.

Capacitance C of a capacitor is defined as the magnitude of the charge Q on the positive plate by the magnitude of the potential difference V between the plates. Therefore, $C=\dfrac{Q}{V}$.

A cylindrical capacitor of length L consists of two cylinders of radii ${{R}_{1}}$ and ${{R}_{2}}$. Let ${{R}_{2}}$> ${{R}_{1}}$.

The outer cylinder is earthed. The cylinder is long enough so that we can neglect the fringing of the electric field at the ends. Electric field at a point between the cylinders will be radial and its magnitude will depend on the distance from the central axis.

Consider a Gaussian surface of length y and radius r such that ${{R}_{2}}$< r < ${{R}_{1}}$.

Flux through the plane surface is zero because electric field and area vectors are perpendicular to each other.

For curved part flux is equal to $\phi =\int{\overrightarrow{E}.d\overrightarrow{s}}=\int{Eds\cos \theta }$, where $\theta $ is the angle between vectors $\overrightarrow{E}$ and $d\overrightarrow{s}$.

Since E and ds are in the same direction, $\theta =0\Rightarrow \cos \theta =\cos 0=1$.

Therefore, $\phi =\int{Eds}$.

Since, E is constant (due to symmetry) on this surface, $\phi =E\int{ds}=EA$ …..(i).

A is the total curved surface area of the cylinder.

We know, $A=2\pi ry$. Substitute the value of A in equation (i).

Therefore, $\phi =EA=E.2\pi ry$ …….(ii).

Let the total charge on the surface of the inner cylinder be Q.

Charge inside the surface is equal to $q=\dfrac{Qy}{L}$.

From Gauss's law, $\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}$ we get,

$\phi =\dfrac{{{Q}_{enclosed}}}{{{\varepsilon }_{\circ }}}=\dfrac{q}{{{\varepsilon }_{\circ }}}=\dfrac{\dfrac{Qy}{L}}{{{\varepsilon }_{\circ }}}=\dfrac{Qy}{L{{\varepsilon }_{\circ }}}$ ……(iii).

From equations (ii) and (iii) we get,

$\dfrac{Qy}{L{{\varepsilon }_{\circ }}}=E.2\pi ry$.

Therefore, $E=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}rL}$ ………(iv).

Potential difference between two points A and B is given by ${{V}_{A}}-{{V}_{B}}=-\int\limits_{{{r}_{B}}}^{{{r}_{A}}}{\overrightarrow{E}.d\overrightarrow{r}}$ .

Here, ${{V}_{A}}$ and ${{V}_{B}}$ are potential of the inner and outer cylinders respectively.

${{r}_{A}}={{R}_{1}}$ and ${{r}_{B}}={{R}_{2}}$.

Therefore,

${{V}_{A}}-{{V}_{B}}=V=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{\overrightarrow{E}.d\overrightarrow{r}}=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{Edr}$. (because E and dr are in the same direction).

Substitute the value of E from equation (iv) in the above equation.

Therefore,

$V=-\int\limits_{{{R}_{2}}}^{{{R}_{1}}}{\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}rl}.dr}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\int\limits_{{{R}_{1}}}^{{{R}_{2}}}{\dfrac{dr}{r}}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\left. \ln r \right|_{{{R}_{1}}}^{{{R}_{2}}}=\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}$.

As discussed above, capacitance $C=\dfrac{Q}{V}=\dfrac{Q}{\dfrac{Q}{2\pi {{\varepsilon }_{\circ }}L}\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}}=\dfrac{2\pi {{\varepsilon }_{\circ }}L}{\ln \dfrac{{{R}_{2}}}{{{R}_{1}}}}$.

Therefore, capacitance of a cylindrical capacitor is directly proportional to its length.

Hence the correct option is A.

Note: In the case of a cylindrical capacitor, charge is stored only on the inner cylinder. The outer cylinder is earthed so it will not have any charge on it. When a plate or conductor is being earth, the potential at that point or surface is considered as zero.

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