A cylindrical air capacitor of length 15.0 m stores 3.20 × 10^-9J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

The given data can be listed below

• The cylinder with length is,$L=15.0\text{m}$.
• The energy stores are,$U=3.20x{10}^{-9}\text{J.}$
• The potential difference of the capacitor is,$V_{ab}=4.00\text{V.}$

Now, we know that the potential energy stored in a charged capacitor is equal to the amount of work done to charge itself. Now to separate opposite charges and place them on different conductorsis given by the following equation:

$$\begin{gathered} \mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{QV}}_{\mathbf{ab}} \\ \mathbf{Q}\mathbf{=}\frac{\mathbf{2}\mathbf{U}}{{\mathbf{V}}_{\mathbf{ab}}} \end{gathered}$$

Now from the above equation we can calculated the charge on the plate

$Q=\frac{2U}{V_{ab}}$

Substitute the $3.2\times {10}^{-9}\text{J}$for U, and 4 V for V_ab in the above equation.

$$\begin{gathered} =\frac{2(3.20\times {10}^{-9}\text{J})}{4.00\text{V}} \\ =1.60\times {10}^{-9}\text{C} \\ =1.60\text{nC} \end{gathered}$$

Now the charge is same for both conductors but is in opposite sign.

Now to find the ratio between the radii of the conductors we will first find the relation between the capacitance and the radii of the conductors:

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

Substitute the 1.6 nC for Q and 4 V for V in the above equation.

$$\begin{gathered} \frac{1.60\times {10}^{-9}\text{C}}{4.00\text{V}} \\ =0.4\times {10}^{-9}\text{F} \end{gathered}$$

Now, the capacitance is related to the radii of the conductors by:

$$\begin{gathered} \frac{C}{L}=\frac{2\pi {\epsilon }_{\circ }}{\ln \left(\frac{r_b}{r_a}\right)} \\ \ln \left(\frac{r_b}{r_a}\right)=\left(\frac{2\pi {\epsilon }_{\circ }}{\frac{c}{L}}\right) \\ \frac{r_b}{r_a}=e^{\left(\frac{2\pi {\epsilon }_{\circ }}{C/L}\right)} \end{gathered}$$

Substitute$0.4\times {10}^{-9}\text{F}$ the for C, 15 m for L in the above equation.

$$\begin{gathered} e^{\left(\frac{2\pi 8.854\times {10}^{-12}\text{F/m}}{0.4\times {10}^{-9}\text{F}/15\text{m}}\right)} \\ =8.05 \end{gathered}$$

Therefore, the radius of the outer conductor is$r_b=8.05r_a.$