Question

A uniform surface charge density of \(20 \mathrm{nC} / \mathrm{m}^{2}\) is present on the spherical surface \(r=0.6 \mathrm{~cm}\) in free space. \((a)\) Find the absolute potential at \(P\left(r=1 \mathrm{~cm}, \theta=25^{\circ}, \phi=50^{\circ}\right) \cdot(b)\) Find \(V_{A B}\), given points \(A(r=2 \mathrm{~cm},\), \(\theta=30^{\circ}, \phi=60^{\circ}\) ) and \(B\left(r=3 \mathrm{~cm}, \theta=45^{\circ}, \phi=90^{\circ}\right)\).

Short answer

Question (a): Calculate the absolute potential at point P in spherical coordinates (1e-2, 60°, 30°) due to a spherical surface with uniform charge density of 20 nC/m² and a radius of 0.6e-2 m. Question (b): Find the potential difference between points A (2e-2, 40°, 20°) and B (3e-2, 80°, 40°) due to the same spherical surface charged.

Step-by-step solution

Find the total charge on the spherical surface

Since the surface charge density is uniform, we can find the total charge on the spherical surface by multiplying the surface charge density with the surface area of the sphere. The surface area of a sphere is given by: \(A = 4 \pi r^2 \) where r is the radius of the sphere. Given: Surface charge density: \(\sigma = 20 \times 10^{-9} \ \mathrm{C/m^2}\) Radius: \(r = 0.6 \times 10^{-2} \ \mathrm{m}\) Total charge on the surface: \(Q = \sigma A = \sigma \cdot 4 \pi r^2 \) Now, let's calculate the total charge on the spherical surface.

Calculate the electric potential due to a spherical surface

The electric potential at any point due to a spherical surface is given by the formula: \(V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}\) where \(V\) is the electric potential, \(Q\) is the total charge on the sphere, \(r\) is the distance from the center of the sphere to the point, and \(\varepsilon_0\) is the vacuum permittivity (\(8.854 \times 10^{-12} \ \mathrm{F/m}\)). Now, let's calculate the electric potential at the given points using the information obtained from step 1.

Find the electric potential at point P (part a)

To find the electric potential at point P: \(V_P = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r_P}\) where \(r_P = 1 \times 10^{-2} \ \mathrm{m}\). After calculating the electric potential at point P, we will move on to finding the electric potential at points A and B.

Find the electric potential at points A and B (part b)

To find the electric potential at point A: \(V_A = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r_A}\) where \(r_A = 2 \times 10^{-2} \ \mathrm{m}\). To find the electric potential at point B: \(V_B = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r_B}\) where \(r_B = 3 \times 10^{-2} \ \mathrm{m}\). Now, we can determine the potential difference between point A and point B.

Find the potential difference between points A and B (part b)

The potential difference between points A and B is given by: \(V_{AB} = V_A - V_B\) Now that we have all the required values, we can find the absolute potential at point P and the potential difference between points A and B by plugging in the values obtained from Steps 1-4 and solving for the electric potential and potential difference.

Key concepts

Surface Charge Density

In electromagnetics, surface charge density is a measure of how much electric charge is accumulated on a surface. It is defined as the amount of charge per unit area and is typically represented by the symbol \( \sigma \). Understanding surface charge density is crucial when dealing with electrostatics problems, as it helps to determine the electric field and potential around charged surfaces.

For instance, in the case of a uniform surface charge density on a spherical surface, as given in our exercise, every point on the sphere's surface has the same amount of charge per unit area. This uniformity simplifies our calculations, as we can easily find the total charge on the sphere by multiplying the charge density by the sphere's surface area. The formula for the surface area of a sphere is \( A = 4 \pi r^2 \), and in our problem, \( \sigma = 20 \times 10^{-9} \ \mathrm{C/m^2} \) and \( r = 0.6 \times 10^{-2} \ \mathrm{m} \). Computing the total charge involves using these values to calculate \( Q = \sigma A \).

Spherical Surface Charge

Dealing with a spherical surface charge, it is essential to appreciate that the symmetry of a sphere greatly influences the resulting electric field and potential. For a sphere with a uniform charge distribution, the electric field outside the sphere is equivalent to that of a point charge located at the center of the sphere, with the value of the charge being equal to the total charge on the sphere.

In our exercise, after finding the total charge on the spherical surface, we apply these principles to determine electric potential at various points around the sphere. The electric potential is a scalar quantity, which means it does not have a direction but rather represents the potential energy per unit charge at a specific point in space due to the presence of electric charges.

Electric Potential Difference

Electric potential difference, also known as voltage, is essentially the work done to move a unit charge from one point to another in an electric field. This concept is crucial when analyzing circuits and electrostatics scenarios because it drives the movement of charge carriers, leading to current.

In our example, we're interested in calculating the potential difference between two points, \( A \) and \( B \) around a charged sphere. We first determine the absolute electric potential at both points using the charges calculated previously. The potential difference, \( V_{AB} = V_A - V_B \), is then the subtraction of potential at point \( A \) from the potential at point \( B \). This difference will indicate how much work is required to move a charge from point \( A \) to point \( B \), which is often a critical step in understanding how charges distribute themselves in space.

Vacuum Permittivity

Vacuum permittivity, denoted as \( \varepsilon_0 \), is a fundamental physical constant that represents the capability of a vacuum to permit the passage of electric field lines. It is also known as the electric constant or permittivity of free space and has a value of approximately \( 8.854 \times 10^{-12} \ \mathrm{F/m} \).

This constant is a part of Coulomb's law, relating to the electric force between two point charges, and it's essential when calculating the electric potential in our problem, particularly in the formula \( V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \). Understanding \( \varepsilon_0 \) is fundamental in electromagnetics as it connects the electric field in a vacuum to the charge and potential. It plays a critical role in our calculations, ensuring that we're considering the properties of the space in which the electrical phenomena occur.