Question

A uniformly charged disk of radius R=35.0 cm carries a charge density of =7.90✕ 10-3 C/m2 . (a) Derive the expression to determine the electric field E on the axis of the disk at the distance x from the center of the disk

Answer 1

We can start considering the surface charge density which is

$\sigma = \frac{Q}{\pi R^2}$

Here,

Q = Total charge and R is the radius of the Disk.

A ring of thickness 'da' centered on the disk as shown has differential area given by

$dA =(2\pi a)da$

And thus a charge given by

$dq = \sigma dA = (2\pi a\sigma) da$

The field produced by this ring of charge is along the x-axis and is given as

$dE_x = \frac{kx2\pi a \sigma da}{(x^2+a^2)^{3/2}} \rightarrow \text{Electric field on the axis of a ring of charge}$

The total field is given by simply integrating over a from 0 to r, then

$E_x = \int^r_0 \frac{(kx2\pi a \sigma )da}{(x^2+a^2)^{3/2}}$

$E_x = kx\pi \sigma \int_0^r \frac{(2a)da}{(x^2+a^2)^{3/2}}$

$E_x = kx\pi \sigma \bigg [ -\frac{1}{2(x^2+a^2)^{1/2}} \bigg]^r_0$

Replacing we have finally,

$E_x = 2\pi k\sigma (1-\frac{x}{(x^2+r^2)^{1/2}})$

Now that is the electric field of a uniformly charged disk:

$E = 2\pi k\sigma (1-\frac{x}{\sqrt{x^2+r^2}})$

Here,

k = Coulomb's constant

$\sigma$ = Surface charge density

x = Distance from center

r = Radius of the Disk

Although the distances are not mentioned in the statement we can establish distances to the center of the disk of approximately 5 cm (In such case the expression could simply be left in terms of x)

Replacing then we would have to,

$E = 2\pi (9*10^9)(7.9*10^{-3})(1-\frac{x}{\sqrt{(x)^2+(35*10^{-2})^2}}})$

Assuming the value of x, we will have,

$E = 2\pi (9*10^9)(7.9*10^{-3})(1-\frac{5*10^{-2}}{\sqrt{(5*10^{-2})^2+(35*10^{-2})^2}}})$

$E = 383.556*10^6N/C$