Question

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer 1

Step 1: Given data

Side of cube$=b$

Magnitude of charge at each vertex$=q$

Step 2: Assumptions

Distance between the centre of the cube and any vertices of the cube $=r$

Length of the diagonal of the cube face $=d$

Length of the diagonal of the cube $=l$

Electric potential due to a single point charge $=v$

Electric potential due to eight charges $=V$

Permittivity of the free space $={\epsilon }_o$

Step 3: Calculation of the electric potential at the centre of the cube

1. Calculation of the distance ($r$) between the centre of the cube and one of the vertices

From the geometry of the cube, it is clear that

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

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

$d=b\sqrt{2}$ ………………(a)

From the geometry of the cube, it is clear that

$l=\sqrt{d^2+b^2}$ ………….(b)

Substituting equation (a) in the equation (b), we get

$l=\sqrt{{\left(\sqrt{2}b\right)}^2+b^2}$

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

$l=\sqrt{3b^2}$

$l=b\sqrt{3}$ ……………(c)

The distance between the centre of the cube and one of the vertices will be equal to half the length of the diagonal of the cube.

$r=\frac{l}{2}$ ……………..(d)

Using equation (c) in equation (d), we get

$r=\frac{b\sqrt{3}}{2}$ ………………(e)

2. Applying the basic equation of electrical potential due to point charge to calculate the net electric potential at the centre of the cube

The electric potential due to point charge is given by the relation

$v=\frac{q}{4\pi {\epsilon }_{o}r}$

Since there are a total of eight charges, therefore the electric potential due to eight charges can be expressed as

$V=\frac{8q}{4\pi {\epsilon }_{o}r}$………………(f)

Using equation (d) in the equation (f), we get

$V=\frac{8q}{4\pi {\epsilon }_o\left({\displaystyle \frac{b\sqrt{3}}{2}}\right)}$

$V=\frac{4q}{\sqrt{3}\pi {\epsilon }_{o}b}$

The above equation gives the electric potential at the centre of the cube due to eight-point charges present at its vertices.

Step 4: Calculation of the electric field at the centre of the cube due to eight-point charges present at its vertices

1. The electric field can be assumed to be made up of hypothetical lines called electric field lines.
2. The interaction between the electric field lines coming from two-point charges determines the magnitude and direction of the electric field at a point.
3. All the point charges placed on the vertices of the cube are equal in magnitude as well as nature i.e they are all either negative charges or positive charges.
4. Let's assume that all the charges on the vertices are negative and let's visualise the interaction of the electric field lines coming from two like charges placed on opposite vertices.
5. Below is the diagram showing the interaction of the electric field lines of the two-point charges placed opposite each other.

6. From figure (a), it is clear that the electric field lines of the two “like point charges” placed oppositely get diverged and cancel their effect at the centre.

7. Presence of no electrical field line at the centre of figure (a) indicates, that there is no electric field at the centre.

8. Similarly the electric field at the centre of the cube due to the like charges placed on the opposite vertices of the cube gets cancelled.

9. If the charges were unlike as depicted by figure (b), then the electric field lines would have converged and the electric field wouldn't have been zero.

Hence, the electric field at the centre of the cube due to like point charges placed on the vertices of the cube is equal to zero.