Question

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.9×109 electrons from one disk to the other causes the electric field strength between them to be 2.0×105 N/C. What are the diameters of the disks?

Answer 1

Concepts and reason

The concepts required to solve the question are charge quantization and the electric field of parallel plate capacitor.

Initially, find the value of charge transferred from one disk to another using the equation: $q = ne$. Then, find the cross-section area of the disk and substitute in the equation of the electric field due to the parallel plate capacitor. Finally, solve the equation for the value of diameter of the disk.

Fundamentals

The charge transfer from one disk to another is given by,

$q = ne$

Here, q is the charge, n is the number of electrons, and e is the charge of electron.

The electric field due to the parallel plate capacitor is given by,

$E = \frac{q}{{\varepsilon A}}$

Here, E is the electric field, q is the charge, A is the cross-section area of the disk, and $\varepsilon$ is the permittivity of free space.

Area of the cross section of the circular disk is,

$A = \pi \frac{{{d^2}}}{4}$

Here, d is the diameter of circular disk.

The expression for the charge transfer from one disk to another is given by the charge quantization equation which is equal to,

$q = ne$

Substitute $2.9 \times {10^9}{\rm{electrons}}$ for n and $\;1.6 \times 10{\;^{ - 19}}{\rm{C}}$ for e.

$\begin{array}{c}\\q = \left( {2.9 \times {{10}^9}} \right)\;\left( {1.6 \times 10{\;^{ - 19}}{\rm{C}}} \right)\\\\ = 4.64 \times 10{\;^{ - 10}}\,{\rm{C}}\\\end{array}$

The charge transfer is $4.64 \times 10{\;^{ - 10}}\,{\rm{C}}$.

The electric field due to the parallel plate capacitor is given by,

$E = \frac{q}{{\varepsilon A}}$

Substitute $\pi \frac{{{d^2}}}{4}$ for A.

$E = \frac{q}{{\varepsilon \pi \frac{{{d^2}}}{4}}}$

Solve the above equation for d.

$\begin{array}{c}\\{d^2} = \frac{q}{{\varepsilon \pi \frac{E}{4}}}\\\\d = \sqrt {\frac{{4q}}{{\varepsilon \pi E}}} \\\end{array}$

Substitute $4.64 \times 10{\;^{ - 10}}\,{\rm{C}}$ for q, $2.0 \times {10^5}{\rm{ N/C}}$ for E, and $\;8.85 \times 10{\;^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {\rm{m}}{{\rm{\;}}^{\rm{2}}}$ for $\varepsilon$ in the equation of electric field due to the parallel plate capacitor.

$d = \sqrt {\frac{{4\left( {4.64 \times 10{\;^{ - 10}}\,{\rm{C}}} \right)}}{{\;\left( {8.85 \times 10{\;^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {\rm{m}}{{\rm{\;}}^{\rm{2}}}} \right)\left( \pi \right)2.0 \times {{10}^5}{\rm{ N/C}}}}}$

$d = \sqrt {3.34 \times 10{\;^{ - 4}}\;{{\rm{m}}^2}}$

Take the square root of d to find the value of diameter.

$d = \pm 1.83 \times {10^{ - 2}}\;{\rm{m}}$

Here, the negative value is ignored because the distance cannot be negative.

Thus, diameter of the disks is $d = 1.83 \times {10^{ - 2}}\;{\rm{m}}$.

Ans:

The diameter of the disks is $d = 1.83 \times {10^{ - 2}}\;{\rm{m}}$.