Question

The potential around a charged sphere. You have an evenly charged spherical surface with radius \(a\) and a total charge \(q\) in a medium with dielectric constant \(D\). Derive the potential inside and outside the surface by using Coulomb's law and integrating over the charged surface.

Short answer

Outside: \(V(r) = \frac{q}{4\pi D r}\), Inside: \(V = \frac{q}{4\pi D a}\)

Step-by-step solution

Understanding Coulomb's Law

Coulomb's law states that the electric potential due to a point charge is given by \[ V = \frac{kq}{r} \] where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge.

Electric Potential Outside the Sphere

To find the potential outside the sphere, consider the sphere as a point charge located at its center. The potential at a distance \(r\) (where \(r \ge a\)) from the center is: \[ V(r) = \frac{q}{4 \pi D r} \] Here, \(D\) is the dielectric constant.

Electric Potential Inside the Sphere

For the potential inside the sphere, we use the fact that an evenly charged surface creates no electric field inside the sphere. Thus, the potential inside remains constant and equal to the potential at the surface: \[ V_{\text{inside}} = V(a) = \frac{q}{4 \pi D a} \] This means that for any point inside the sphere (where \(r < a\)), the potential is constant and equal to the potential at radius \(a\).

Integrating Over the Charged Surface

The results obtained from Step 2 and Step 3 are verified by integrating Coulomb's Law over the charged spherical surface. Note that for a uniformly charged surface, this integral supports our earlier findings.

Key concepts

Coulomb's law

Coulomb's law is a fundamental principle in electrostatics that describes the electric force between two point charges. It states that the electric potential, or voltage, due to a single point charge is given by

• \( V = \frac{kq}{r} \)

\k\ is the Coulomb's constant, \q\ is the charge, and \r\ is the distance from the charge. When dealing with a charged sphere, we can use Coulomb’s law to find the potential at various points in space. For points outside the sphere, we treat the entire sphere as if all its charge were concentrated at the center. For points inside, the field is determined by integrating over the charged surface.

dielectric constant

The dielectric constant, or relative permittivity, \( D \) , is a measure of how much a material can reduce the electric field compared to a vacuum. When a dielectric medium surrounds a charged object, the potential at a point is given by

• \( V = \frac{q}{4 \pi D r} \)

This formula shows that the potential decreases with increasing \( D \). Materials with higher dielectric constants are better at insulating against electric fields. In our problem, knowing \( D \) allows us to calculate the potential around the charged sphere accurately.

electric field in charged surfaces

For a charged sphere, understanding the electric field is crucial. Outside the sphere, the electric field behaves as if all the charge were concentrated at the center. Inside an evenly charged spherical surface, the electric field is zero, and this results in a constant potential. Therefore, the potential at a distance \( r \) from the center (outside the surface) is

• \( V = \frac{q}{4 \pi D r} \)

For points inside the sphere, since there is no electric field, the potential remains constant at

• \( V = \frac{q}{4 \pi D a} \)

where \( a \) is the radius of the sphere.

electrostatics integration

Integration in electrostatics requires summing contributions of the electric potential from each infinitesimal charge element on the charged surface. To find the total potential at a point outside or on a charged sphere, we calculate the integral of Coulomb's law over the sphere's surface. This verifies the formulas for the potential derived for inside and outside the sphere.

uniform charge distribution

A uniform charge distribution means the charge is spread evenly across the surface or volume of an object. For a spherical surface with radius \( a \) and total charge \( q \), the charge is evenly spread out. This simplifies calculations of electric potential and field. Inside the sphere, the potential is constant because the field cancels out. Outside, the potential decreases with distance, as calculated using the formula

• \( V(r) = \frac{q}{4 \pi D r} \)

for \( r \ge a \) .