Questions: Exercise 11: Consider a uniformly charged disk of radius R and charge density σ, with a total charge Q lying in the xy-plane. 1- Find the electric field at a point P, located at a distance z, along the z-axis that passes through the center of the disk perpendicular to its plane. 2- What is the electric potential at a distance z from the axis? 3- From the electric potential, show that the corresponding electric field at P is exactly the result obtained early from Coulomb's law.

Solution Steps

1. Consider a small ring element of radius \( r \) and thickness \( dr \) on the disk.
2. The charge \( dq \) on this ring is given by \( dq = \sigma \cdot 2\pi r \, dr \).
3. The electric field \( dE \) due to this ring at point \( P \) is directed along the \( z \)-axis because of symmetry.
4. The contribution to the electric field from this ring is \( dE = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq \cdot z}{(r^2 + z^2)^{3/2}} \).
5. Integrate \( dE \) from \( r = 0 \) to \( r = R \) to find the total electric field \( E \) at point \( P \).

1. The potential \( dV \) due to the ring element is \( dV = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq}{\sqrt{r^2 + z^2}} \).
2. Integrate \( dV \) from \( r = 0 \) to \( r = R \) to find the total potential \( V \) at distance \( z \).

1. The electric field \( E \) is related to the potential \( V \) by \( E = -\frac{dV}{dz} \).
2. Differentiate the expression for \( V \) with respect to \( z \) to find \( E \).
3. Show that this expression for \( E \) matches the result obtained from the direct calculation using Coulomb's law.

Final Answer