Question

A sphere of radius \(1 \mathrm{~cm}\) has potential of \(8000 \mathrm{v}\), then energy density near its surface will be ...... (A) \(64 \times 10^{5}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (C) \(8 \times 10^{3}\left(\mathrm{~J} / \mathrm{m}^{3}\right)\) (D) \(32\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

Short answer

The short answer is: \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\).

Step-by-step solution

Calculate the electric field near the surface

To find the electric field (E) near the surface of the sphere, we can use the formula: E = \( \frac{V}{r} \), where V is the potential, and r is the radius. Given, V = 8000 V, and r = \(1 \times 10^{-2}\) m (1 cm is converted to meters) Now, let's calculate E: E = \( \frac{8000}{1 \times 10^{-2}} \) = \(8 \times 10^{5} \frac{V}{m} \)

Calculate the energy density near the surface

Now that we have the value for the electric field (E), we can find the energy density (u) using the formula: u = \( \frac{1}{2}ε_{0}E^{2} \), where \(ε_{0}\) is the vacuum permittivity (\( 8.854 \times 10^{-12} \frac{C^2}{N.m^2} \)) Let's calculate u: u = \( \frac{1}{2}(8.854 \times 10^{-12})(8 \times 10^{5})^{2} \) = \(2.83 \times 10^{6} \frac{J}{m^3} \) After calculating the energy density, we can see that the correct option is: (B) \(2.83\left(\mathrm{~J} / \mathrm{m}^{3}\right)\)

Key concepts

Electric Field

The electric field is an essential concept in electrostatics. It's a vector field around a charged object and describes the force that would act on a charge placed in this field. For a sphere, the electric field near its surface can be computed using the formula:\[ E = \frac{V}{r} \]where:

• \( V \) is the potential, measured in volts (V).
• \( r \) is the radius of the sphere, converted to meters.

In the given problem, we have a potential \( V = 8000 \) volts and a sphere radius \( r = 1 \times 10^{-2} \) meters, as 1 cm needs to be converted to meters. By substituting these values into the formula, we get:\[ E = \frac{8000}{1 \times 10^{-2}} = 8 \times 10^{5} \text{ V/m} \]This calculation gives us the intensity of the electric field at the sphere's surface.

Energy Density

Energy density is a measure of the amount of energy stored in a given system or region of space per unit volume. In the context of electric fields, it provides insight into how much energy is present around the charged sphere. The formula for the energy density \( u \) of an electric field is:\[ u = \frac{1}{2} \varepsilon_{0} E^2 \]where:

• \( \varepsilon_{0} \) is the vacuum permittivity.
• \( E \) is the electric field strength.

Using our previously calculated electric field \( E = 8 \times 10^{5} \text{ V/m} \), and the value for vacuum permittivity \( \varepsilon_{0} = 8.854 \times 10^{-12} \text{ F/m} \), the energy density is:\[ u = \frac{1}{2} (8.854 \times 10^{-12}) (8 \times 10^{5})^2 = 2.83 \times 10^{6} \text{ J/m}^3 \]This indicates the energy concentration around the sphere is quite significant.

Vacuum Permittivity

Vacuum permittivity, denoted as \( \varepsilon_{0} \), is a fundamental constant in electromagnetism. It describes how an electric field affects and is affected by a vacuum. The value of vacuum permittivity is \( 8.854 \times 10^{-12} \text{ C}^2/\text{N.m}^2 \). It is crucial in equations like:

• Calculating the energy density, where \( \varepsilon_{0} \) determines the energy storage capability in vacuum.
• Describing the force between point charges through Coulomb's law.

In our problem, \( \varepsilon_{0} \) directly impacts the computed energy density, showing the relation between the field created by charges and stored energy in vacuum.

Sphere

Spheres in electrostatics serve as perfect models for studying electric fields and potentials due to their symmetrical properties. The electric field of a charged sphere behaves like that of point charges but allows analytical solutions of equations:

• The field outside a uniformly charged sphere is equivalent to having all the charge concentrated at its center.
• Inside, the electric field is uniform, making calculations applicable to practical realities, like conducting spheres.

By employing simple geometric principles and formulas, understanding such problems becomes intuitive, allowing one to compute values like potential and energy density effectively. The sphere in our exercise exemplifies this by simplifying evaluations of electric field and energy concentration near its surface.