Question

Two closely spaced circular disks form a parallel-plate capacitor. Transferring 2,632,537,442.25 electrons from one disk to the other causes the electric field strength to be 1,832,461.64 N/C. What are the diameters, in mm, of the disks?

Answer 1

Given:

Amount of electrons = 2,632,537,442.25

Electric field strength, E = 1,832,461.64 N/C

Let's find the diameters.

Apply the formula:

$\begin{gathered} E=\frac{v}{d} \\ v=E*d \end{gathered}$

The charge on the capacitor will be expressed by:

$\begin{gathered} Q=CV \\ \\ \text{ Where:} \\ C=\frac{fA}{d} \\ \\ \text{ Thus, we have:} \\ Q=\frac{fA}{d}*E*d \end{gathered}$

A is the area.

We have:

$\begin{gathered} A=\pi r^2 \\ \\ E=\frac{Q}{\pi r^2*\epsilon_o} \\ \\ r=\sqrt{\frac{Q}{E\pi *\epsilon_o}} \end{gathered}$

Now, let's solve for the radius, r

Plug in values and solve for r

$r=\sqrt{\frac{2632537442.25*1.6*10^{-19}}{1832461.64*\pi *8.854*10^{-12}}}$

SOlving further:

$r=0.00287\text{ m}$

Also, we know:

Diameter = radius x 3

Diameter = 0.00287 x 2

Diameter = 0.00574 m

The diameter in mm will be = 5.74 mm

ANSWER:

5.74 mm