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=(v)/(d) \\ v=E*d \end{gathered}`

The charge on the capacitor will be expressed by:

`\begin{gathered} Q=CV \\ \\ \text{ Where:} \\ C=(fA)/(d) \\ \\ \text{ Thus, we have:} \\ Q=(fA)/(d)*E*d \end{gathered}`

A is the area.

We have:

`\begin{gathered} A=\pi r^2 \\ \\ E=(Q)/(\pi r^2*\epsilon_o) \\ \\ r=\sqrt{(Q)/(E\pi *\epsilon_o)} \end{gathered}`

Now, let's solve for the radius, r

Plug in values and solve for r

`r=\sqrt{(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