Question 23.3: The Electric Field of a Uniformly Charged Disk A disk of rad...
The Electric Field of a Uniformly Charged Disk
A disk of radius R has a uniform surface charge density σ. Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk (Fig. 23.4).
Conceptualize If the disk is considered to be a set of concentric rings, we can use our result from Example 23.2— which gives the field created by a single ring of radius a— and sum the contributions of all rings making up the disk. By symmetry, the field at an axial point must be along the central axis.
Categorize Because the disk is continuous, we are evaluating the field due to a continuous charge distribution rather than a group of individual charges.
Analyze Find the amount of charge dq on the surface area of a ring of radius r and width dr as shown in Figure 23.4:
d q =\sigma d A=\sigma(2 \pi r d r)=2 \pi \sigma r d rUse this result in Equation (3) in Example 23.2 (with a replaced by r and Q replaced by dq) to find the field due to the ring:
d E_x =\frac{k_e x}{\left(r^2+x^2\right)^{3 / 2}}(2 \pi \sigma r d r)To obtain the total field at P, integrate this expression over the limits r = 0 to r = R, noting that x is a constant in this situation:
\begin{aligned}E_x & =k_e x \pi \sigma \int_0^R \frac{2 r d r}{\left(r^2+x^2\right)^{3 / 2}} \\& =k_e x \pi \sigma \int_0^R\left(r^2+x^2\right)^{-3 / 2} d\left(r^2\right) \\& =k_e x \pi \sigma\left[\frac{\left(r^2+x^2\right)^{-1 / 2}}{-1 / 2}\right]_0^R=2 \pi k_e \sigma\left[1-\frac{x}{\left(R^2+x^2\right)^{1 / 2}}\right]\end{aligned}Finalize This result is valid for all values of x > 0. For large values of x, the result above can be evaluated by a series expansion and shown to be equivalent to the electric field of a point charge Q. We can calculate the field close to the disk along the axis by assuming x << R; in this case, the expression in brackets reduces to unity to give us the near-field approximation
E=2 \pi k_e \sigma=\frac{\sigma}{2 \epsilon_0}where \epsilon_0 is the permittivity of free space.