A cylindrical capacitor has radii aand bas in Fig. 25-6. Show that half thestored electric potential energy lieswithin a cylinder whose radius is$\mathit{R}\mathbf{=}\sqrt{\mathbf{a}\mathbf{b}}$.

• Radius of the inner cylinder is
• Radius of the outer cylinder is b
• Charge on the surface of the inner cylinder is positive
• Charge on the outer cylinder is negative.

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

The line charge density is equal to the electric charge per unit length.

We use the formula of the electric field outside the cylinder. Using the line charge density formula, we modify the electric field. And finally putting this value in energy density and integrating it we the required result.

Formulae:

The energy density of a capacitor plate, $u=\frac{1}{2}{\epsilon }_0{E}^2$ ...(i)

Here, u is the energy density, ${\epsilon }_0$is the permittivity of the free space, and E is electric field.

The electric field due to a line charge, $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ ...(ii)

Here, $\lambda$is line charge density, $r$is the distance from the line charge, and ${\epsilon }_0$ is the permittivity of the free space.

The line charge density of a distribution, $\lambda =\frac{q}{L}$ ...(iii)

Here, $\lambda$ is line charge density, $q$ is the electric charge, $L$is the length.

The energy stored within the plates, $U_R=\int udv$ ...(iv)

$U_R$is the energy stored, u is the energy density,dv is a very small volume.

We first find the energy stored in cylinder of radius R and length L, whose surface lies between the inner and outer cylinders of capacitor$\left(a<R<b\right)$.

Now, the value of the electric field due to a line charge can be given using equation (iii) in equation (ii) as follows:

$E=\frac{q}{2\pi {\epsilon }_{0}rL}$

And the energy density using the above value in equation (i) at that point can be given as follows:

$$\begin{gathered} u={\epsilon }_0\frac{q^2}{8{\pi }^2{\epsilon }_0^2{r}^2{L}^2} \\ =\frac{q^2}{8{\pi }^2{\epsilon }_0{r}^2{L}^2} \end{gathered}$$

Now, the energy stored using the above value in equation (iv)is given by the volume integral as follows: ( $dv=2\pi rLdr$ for cylinder of length L .)

$$\begin{gathered} U_R={\int }_a^R\frac{q^2}{8{\pi }^2{\epsilon }_0{r}^2{L}^2}2\pi rLdr \\ =\frac{q^2}{4\pi {\epsilon }_0{L}_a}{\int }_a^R\frac{dr}{r} \\ =\frac{q^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{L}}In\left(\frac{R}{a}\right) \end{gathered}$$ …(v)

The total energy stored in the capacitor R by b can be given as:

$U_b=\frac{q^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{L}}In\left(\frac{b}{a}\right)$ …(vi)

Now, for the ration of above two equations to be halved, we can get the radius values as:

$$\begin{gathered} In\left(\frac{R}{a}\right)=\frac{1}{2}In\left(\frac{b}{a}\right) \\ In\left(\frac{R}{a}\right)=In{\left(\frac{b}{a}\right)}^{\frac{1}{2}} \\ In\left(\frac{R}{a}\right)=In\sqrt{\left(\frac{b}{a}\right)} \\ \left(\frac{R}{a}\right)=\sqrt{\left(\frac{b}{a}\right)} \\ R=\sqrt{\frac{b}{a}{a}^2} \\ =\sqrt{ab} \end{gathered}$$

Hence, the value of the radius is $\sqrt{ab}$.