FANDOM


Problem

Find the electric field a distance $ x $ along the axis from a disc of radius $ R $ and uniform charge density $ \sigma $.
Disk
Hint: a disk can be thought of as a bunch of concentric rings. Begin by finding the electric field a distance $ x $ along the axis up from a thin ring of charge $ dQ $ and radius $ a $.

You may use the integral

$ \int \frac{xdx}{(\sqrt{x^2 + a^2})^3} = -\frac{1}{\sqrt{x^2 + a^2}} + C $.

Solution

Ring

Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is $ r = \sqrt{x^2 + a^2} $.

Hence, the differential element of electric field (along the x-axis) is

$ dE_x = dE \cos{(\theta)} $
$ dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{dQ}{x^2 + a^2} \frac{x}{\sqrt{x^2 + a^2}} $.

Integrating yields,

$ E_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{x}{(\sqrt{x^2 + a^2})^3} \int_{0}^{Q} dQ $
$ E = \frac{1}{4\pi {\epsilon}_{0}} \frac{xQ}{(\sqrt{x^2 + a^2})^3} \boldsymbol{i} $.

For the disk problem, replace the differential charge with $ dQ = \sigma 2\pi r dr $ because the areal charge density is defined as $ \sigma = \frac{dQ}{dA} $ and $ A = \pi r^2 $ for a disk.

Therefore, the electric field is modified to (we are integrating with respect to the variable $ r $ now, so $ a $ is replaced with $ r $)

$ dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{\sigma 2\pi rdr}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}} $.

Using the given integral yields,

$ E_x = \frac{1}{4\pi {\epsilon}_{0}} \int_{0}^{R} \frac{(\sigma 2\pi rdr) x}{(\sqrt{x^2 + r^2})^3} $
$ E_x = \frac{\sigma x}{2 {\epsilon}_{0}} \left( -\frac{1}{\sqrt{x^2 + r^2}} + \frac{1}{x} \right) $.

Therefore,

$ E = \frac{\sigma}{2 {\epsilon}_{0}} \left(1 -\frac{x}{\sqrt{x^2 + r^2}} \right) \boldsymbol{i} $.
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