Question

The space between two concentric conducting spherical shells of radii b = 1.70 cmand a = 1.20 cmis filled with a substance of dielectric constant $\mathit{\kappa }\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{5}$. A potential difference V = 73.0Vis applied across the inner and outer shells.(a)Determine the capacitance of the device(b)Determine the free charge q on the inner shell(c)Determine the charge q'induced along the surface of the inner shell.

Short answer

a) The capacitance of the device is 0.107 nF .

b) The free charge on the inner shell is 7.81 nC .

c) The charge induced along the surface of the inner shell is 7.47nC .

Step-by-step solution

The given data

• Outer radius of the shell, b = 1.70 cm
• Inner radius of the shell, a = 1.20 cm
• Dielectric constant of the substance,$\kappa =23.5$
• Potential difference across the inner and outer shells, V=73.0V

Understanding the concept of the free charge and capacitance

A capacitor consists of two isolated plates with charges and. The capacitanceof the capacitor is defined from,

Here,is the potential difference between the two plates.

By using the given equation, we can find the capacitance of the device. Also by using the relation between charge and capacitance, we can find the charge on the inner shell. By using the equation, we can find the induced charge on the inner shell.

Formulae:

The charge stored between the capacitors, q = CV …(i)

Here, q is the electric charge, C is the capacitance, and V is the potential difference between the two plates.

The induced dielectric charge on each plate,$q-q\text{'}=\frac{q}{\kappa }$ ...(ii)

Here, q is the electric charge, q' is the induced dielectric charge,$\kappa$is the dielectric constant.

The capacitance of a spherical capacitor, $C_0=4{\mathrm{\pi \epsilon }}_0\mathrm{\kappa }\left(\frac{\mathrm{ab}}{\mathrm{b}-\mathrm{a}}\right)$ ...(iii)

Here,$C_0$ is the capacitance,${\epsilon }_0$ is the permittivity of the free space, $\kappa$is the dielectric constant, a and b are radii of the concentric spherical plates.

(a) Calculation of the capacitance of the device

By substituting the given data in equation (iii),we can find the capacitance of the spherical capacitor as

$$\begin{gathered} C_0=4\times 3.14\times 23.5\times \left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\times \left(\frac{1.7\times {10}^{-12}m\times 1.20\times {10}^{-2}m}{1.7\times {10}^{-12}m-1.20\times {10}^{-2}m}\right) \\ =2.61\times {10}^{-9}\times \left(\frac{2.04\times {10}^{-4}}{0.5\times {10}^{-2}}\right) \\ =1.07\times {10}^{-10}F \\ =0.107nF \end{gathered}$$

Hence, the value of the capacitance is 0.107nF .

(b) Calculation of the inner charge on the shell

The charge on the positive plate can be given using equation (i) as follows:

$$\begin{gathered} q=1.07\times {10}^{-10}F\times 73V \\ =7.81\times {10}^{-9}C \\ =7.81nC \end{gathered}$$

Hence, the value of the charge is 7.78nC .

(c) Calculation of the induced charge along the surface of the inner shell

The value of the induced charge on the surface of the inner shell due to the dielectric material can be given using the given data in equation (ii) as:

$$\begin{gathered} q\text{'}=q\left(1-\frac{1}{\kappa }\right) \\ =\left(7.81\times {10}^{-9}C\right)\times \left(1-\frac{1}{23.5}\right) \\ =7.81\times {10}^{-9}C\times 0.957 \\ =7.81\times {10}^{-9}C \\ =7.47nC \end{gathered}$$

Hence, the value of the surface charge is 7.45nC .