Section 3.3: Potential of Continuous Charges
For continuous charge distributions, the electric potential at a point is obtained by integrating the contributions of infinitesimal charge elements over the distribution.
Formula:
\[ V = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r} \]
where \( dq \) is the infinitesimal charge element and \( r \) is the distance from \( dq \) to the point of interest.
Applications:
- Line charge: \( dq = \lambda dx \)
- Surface charge: \( dq = \sigma dA \)
- Volume charge: \( dq = \rho dV \)
Example 1
Compute the potential at a point on the axis of a uniformly charged rod of length L and linear charge density λ.
Take a point at distance z from the center of the rod. Let \( x \) run along the rod:
\[ V = \frac{1}{4\pi\epsilon_0} \int_{-L/2}^{L/2} \frac{\lambda dx}{\sqrt{x^2 + z^2}} = \frac{\lambda}{4\pi\epsilon_0} \ln \frac{L/2 + \sqrt{(L/2)^2 + z^2}}{-L/2 + \sqrt{(L/2)^2 + z^2}} \]
Practice Problems
- Find the potential on the axis of a uniformly charged ring of radius R and total charge Q.
- Compute the potential at the center of a uniformly charged semicircular arc.
- Potential due to a uniformly charged disk at a point along its axis.
- Explain why potential simplifies calculation compared to directly integrating the electric field.
- For a sphere with uniform volume charge density ρ, find the potential at an external point.