Question

Few q charges are placed at the vertices of the cube. find electrostatic force. Determine the potential and electric field due to this charge array at the center of the cube.

Answer 1

Step 1. Given data:

Le a cube of dimensions shown in the figure below:

Refer to the diagram for the respective dimensions mentioned:

Let the sides of the cube = $b$

Given, that the charge at the vertices = $q$

Step 2. Formula used:

The electric potential (V) at the center of the cube is due to the presence of eight charges at the vertices.

$V=\raisebox{1ex}{$8\times {\displaystyle \frac{q}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}}=8q$}\!\left/ \!\raisebox{-1ex}{$4{\mathrm{\pi \epsilon }}_0$}\right.r$

Where $V$= Electric potential. $r$=distance of any of the vertices to the center of the cube.

Step 3. Calculation:

Diagonal of one of the sides of the cube may be given by:

$d^2=\sqrt{b^2+b^2}$= $\sqrt{2b^2}$

$d=b\sqrt{2}$

Length of the diagonal of the cube

$$\begin{gathered} l^2=\sqrt{d^2+b^2}=\sqrt{2b^2+b^2}=\sqrt{3b^2} \\ l=b\sqrt{3} \end{gathered}$$

The distance between any one of the vertices and the center of the cube is

The electric potential ($V$) at the center of the cube is due to the eight charges at the vertices ( by putting the value of $r$)

$$\begin{gathered} V=\raisebox{1ex}{$8q$}\!\left/ \!\raisebox{-1ex}{$4{\mathrm{\pi \epsilon }}_0$}\right.=\raisebox{1ex}{$8q$}\!\left/ \!\raisebox{-1ex}{$4{\mathrm{\pi \epsilon }}_0\left(\mathrm{b}{\displaystyle \frac{\sqrt{3}}{2}}\right)$}\right. \\ =\raisebox{1ex}{$4q$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{b}$}\right. \end{gathered}$$

Therefore, the potential at the center of the cube will be

= $\raisebox{1ex}{$4q$}\!\left/ \!\raisebox{-1ex}{$\left(\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{b}\right)$}\right.$

Step 4. Find electric field intensity:

1. Electric field is zero in the center because the sum of electric field vectors have the same intensity and direction, but are opposite.
2. That point is halfway between two like charges
3. Hence the intensity of the electric field at the center of the cube, due to all the eight charges is zero. The charges are distributed equally with respect to the center of the cube.
4. Hence, they get canceled.

Hence, the potential at the center of the cube is $\raisebox{1ex}{$4q$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{b}$}\right.$ and the electric field intensity at the center is zero.