Potential Due to Point and Continuous Charge Distribution

Potential Due to Point and Continuous Charge Distributions

In electrostatics, the electric potential (or simply potential) is a scalar quantity that describes the potential energy per unit charge at a point in space due to the presence of electric fields. It is closely related to the electric field, as the electric field is the negative gradient of the potential. The electric potential at a point is a measure of the work done by an external force in bringing a positive test charge from infinity to that point without any acceleration.

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1. Electric Potential Due to a Point Charge

The electric potential $V$ due to a point charge $Q$ at a distance $r$ from the charge is given by:

$$V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$

Where:

• $V$ is the electric potential at a distance $r$ from the point charge,
• $Q$ is the magnitude of the point charge,
• $r$ is the distance from the point charge to the point where the potential is being calculated,
• $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{C}^2 / \text{N·m}^2$).

Key Points:

• The potential due to a point charge is a scalar quantity and is radial (it depends only on the distance from the charge, not on direction).
• The potential is positive for a positive charge and negative for a negative charge.
• At infinity, the potential is defined to be zero, so the potential at a distance $r$ is relative to the potential at infinity.

Electric Field and Potential:

• The electric field $E$ and electric potential $V$ are related through the gradient:

$$\vec{E} = - \nabla V$$

For a point charge, the electric field is radially outward (or inward for a negative charge), and we can obtain the electric field from the potential:

$$E = - \frac{dV}{dr} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$$

This matches Coulomb’s law for the electric field due to a point charge.

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2. Electric Potential Due to a Continuous Charge Distribution

For a continuous charge distribution, the potential at a point in space is the sum of the potentials due to all infinitesimal charge elements that make up the distribution. Depending on the geometry of the charge distribution, we can integrate to find the total potential.

The general expression for the potential at a point $\vec{r}$ due to a continuous charge distribution is:

$$V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{|\vec{r} - \vec{r}'|}$$

Where:

• $dq$ is the infinitesimal charge element,
• $\vec{r}$ is the position where the potential is being calculated,
• $\vec{r}'$ is the position of the charge element $dq$,
• $|\vec{r} - \vec{r}'|$ is the distance between the charge element and the point where the potential is being calculated.

The specific form of the integral depends on the geometry of the charge distribution.

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3. Electric Potential Due to a Line Charge Distribution

Consider a uniformly charged infinite line or a line of charge with charge per unit length $\lambda$ (charge density) placed along the x-axis. To find the potential at a point $P$ located a distance $r$ from the line, we use the following approach:

Solution:

• The potential due to an infinitesimal charge element $dq = \lambda dx$ at a point located at a distance $r$ is:

$$dV = \frac{1}{4 \pi \epsilon_0} \frac{dq}{r}$$

• The distance from each charge element to the point $P$ is $r$, which is constant for the line charge (if the line is perpendicular to the point).

For a finite line of charge with total charge $Q$ and length $L$, the potential at a point $P$ located at a perpendicular distance $r$ from the line is:

$$V = \frac{1}{4 \pi \epsilon_0} \int_{-L/2}^{L/2} \frac{\lambda dx}{\sqrt{x^2 + r^2}}$$

This integral can be evaluated to give the potential for a finite line of charge.

For an infinite line of charge (with $L \to \infty$), the potential at a point at a distance $r$ from the line is:

$$V(r) = \frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{1}{r} \right)$$

This shows that the potential due to an infinite line of charge grows logarithmically with distance from the line.

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4. Electric Potential Due to a Surface Charge Distribution

Consider a uniformly charged disk with surface charge density $\sigma$ and radius $R$. To find the potential at a point along the axis of the disk (say, at a distance $z$ from the center), we treat the disk as a collection of infinitesimal charge elements and integrate over the entire surface.

Solution:

• An infinitesimal charge element $dq = \sigma dA = \sigma r \, dr \, d\theta$ is located at a distance $r$ from the center of the disk.
• The potential due to $dq$ at a point on the axis of the disk (distance $z$ from the center) is:

$$dV = \frac{1}{4 \pi \epsilon_0} \frac{dq}{\sqrt{r^2 + z^2}}$$

• The total potential $V$ is obtained by integrating over the entire surface of the disk:

$$V(z) = \frac{1}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{\sigma r \, dr \, d\theta}{\sqrt{r^2 + z^2}}$$

The solution to this integral gives the electric potential at a point on the axis of the disk. After evaluating the integral:

$$V(z) = \frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + z^2} - z \right)$$

This expression gives the potential at a point along the axis of a uniformly charged disk.

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5. Electric Potential Due to a Spherically Symmetric Charge Distribution

For a spherically symmetric charge distribution, such as a uniformly charged sphere, the potential at a point outside the distribution can be derived by considering the charge as if it were concentrated at the center of the sphere.

Solution:

• For a uniformly charged sphere with total charge $Q$ and radius $R$, the electric potential at a point outside the sphere (at a distance $r$ from the center) is:

$$V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$

This is the same as the potential due to a point charge, because from a distance $r$ (where $r > R$), the sphere behaves like a point charge.

• For points inside the sphere (at a distance $r$ where $r < R$), the potential at a point inside the sphere is:

$$V(r) = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q}{R} - \frac{Q}{2R^3} r^2 \right)$$

This expression arises from the fact that the electric potential inside a uniformly charged sphere depends on the distance from the center.

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6. Summary of Potential for Different Charge Distributions

• Point Charge: The potential due to a point charge $Q$ at a distance $r$ is $V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$.
• Line Charge: For a line charge with linear charge density $\lambda$, the potential at a distance $r$ from the line is $V(r) = \frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{1}{r} \right)$ for an infinite line of charge.
• Surface Charge: For a uniformly charged disk, the potential at a point on its axis is $V(z) = \frac{\sigma}{2 \epsilon_0} \left( \sqrt{R^2 + z^2} - z \right)$, where $\sigma$ is the surface charge density.
• Spherical Symmetry: For a spherically symmetric charge distribution, the potential outside the sphere is $V(r) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$ (same as a point charge), and inside the sphere, it follows a quadratic dependence on $r$.

The electric potential plays a crucial role in understanding the behavior of