A cube of side 'a' has a charge Q at each of its vertices. What is the potential due to this charge array at the centre of the cube?

Question

Prepp Team · Accepted Answer

Calculating Electric Potential at the Cube Center

This problem asks us to find the electric potential at the very center of a cube. We are given that the cube has a side length of 'a' and that there is a charge Q located at each of its eight vertices.

Electric potential is a scalar quantity, meaning it doesn't have a direction like electric field. The total potential at a point due to a collection of charges is simply the sum of the potentials due to each individual charge. This is known as the principle of superposition for electric potential.

Key Concepts in Potential Calculation

Electric Potential: The electric potential at a point in an electric field is defined as the amount of work needed per unit of charge to move a test charge from infinity to that point. For a point charge Q, the potential V at a distance r is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$, where $\epsilon_0$ is the permittivity of free space.
Superposition Principle: The total electric potential at a point due to a system of charges is the algebraic sum of the potentials due to each individual charge.

Step-by-Step Solution

Step 1: Determine the Distance from Vertex to Center

To use the potential formula, we first need to find the distance from each vertex (where a charge Q is located) to the exact center of the cube.

Consider a cube with side length 'a'.

The diagonal of one face of the cube can be found using the Pythagorean theorem: $d_{face}^2 = a^2 + a^2 = 2a^2$, so the face diagonal is $d_{face} = a\sqrt{2}$.
The main diagonal (body diagonal) of the cube connects opposite vertices and passes through the center. We can find its length using the Pythagorean theorem again, considering one side of the cube and the face diagonal: $d_{body}^2 = a^2 + d_{face}^2 = a^2 + (a\sqrt{2})^2 = a^2 + 2a^2 = 3a^2$. So, the body diagonal is $d_{body} = a\sqrt{3}$.
The center of the cube is located exactly at the midpoint of each body diagonal. Therefore, the distance 'r' from any vertex to the center is half the length of the body diagonal: $r = \frac{d_{body}}{2} = \frac{a\sqrt{3}}{2}$.

Every one of the eight vertices is located at this same distance 'r' from the center of the cube.

Step 2: Calculate Potential from a Single Charge

Now, let's calculate the electric potential at the center of the cube due to just one of the charges Q located at a vertex. Using the formula $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$ with $r = \frac{a\sqrt{3}}{2}$:

$V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{(a\sqrt{3}/2)}$

$V_1 = \frac{1}{4\pi\epsilon_0} \frac{2Q}{a\sqrt{3}}$

$V_1 = \frac{2Q}{4\pi\epsilon_0 a\sqrt{3}}$

$V_1 = \frac{Q}{2\pi\epsilon_0 a\sqrt{3}}$

Step 3: Calculate Total Potential from All Charges

Since there are 8 identical charges Q, one at each vertex, and all are at the same distance 'r' from the center, we can find the total potential at the center by summing the potentials due to each charge. By the principle of superposition:

$V_{total} = V_1 + V_2 + ... + V_8$

Since $V_1 = V_2 = ... = V_8 = \frac{Q}{2\pi\epsilon_0 a\sqrt{3}}$, the total potential is:

$V_{total} = 8 \times V_1$

$V_{total} = 8 \times \frac{Q}{2\pi\epsilon_0 a\sqrt{3}}$

$V_{total} = \frac{8Q}{2\pi\epsilon_0 a\sqrt{3}}$

$V_{total} = \frac{4Q}{\pi\epsilon_0 a\sqrt{3}}$

This can also be written as $\frac{4Q}{\sqrt{3}\pi\epsilon_0 a}$ by rearranging the terms in the denominator.

Summary of Potential Calculation

The electric potential at the center of a cube with side 'a' and charge Q at each of its 8 vertices is found to be $\frac{4Q}{\sqrt{3}\pi\epsilon_0 a}$.

Let's compare this result with the given options:

Option 1: $\frac{2Q}{\sqrt{3}\pi\epsilon_0 a}$
Option 2: $\frac{4Q}{\sqrt{3}\pi\epsilon_0 a}$
Option 3: $\frac{Q}{2\sqrt{3}\pi\epsilon_0 a}$
Option 4: $\frac{8Q}{\sqrt{3}\pi\epsilon_0 a}$

Our calculated result matches Option 2.

Revision Table: Related Electrostatics Concepts

Review these key concepts related to electric potential and fields:

Concept	Definition	Formula (if applicable)
Electric Potential (V)	Potential energy per unit charge at a point. Scalar quantity.	$V = \frac{U}{q_0}$ or $V = \int \vec{E} \cdot d\vec{l}$
Potential due to Point Charge Q	Potential at a distance r from a source charge Q.	$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
Superposition Principle (Potential)	Total potential at a point from multiple charges is the sum of individual potentials.	$V_{total} = \sum V_i = \sum \frac{1}{4\pi\epsilon_0} \frac{Q_i}{r_i}$
Electric Field (&vec;E)	Force per unit charge at a point. Vector quantity.	$\vec{E} = \frac{\vec{F}}{q_0}$ or $\vec{E} = -\nabla V$

Additional Information: Understanding Electric Potential

Electric potential is a very useful concept in electrostatics because it simplifies calculations compared to working with electric fields, especially for calculating the potential energy of a charge in a field. Unlike electric field, which is a vector and requires dealing with components and directions, electric potential is a scalar, so we just add up the contributions from all charges algebraically.

In symmetric charge distributions, like the cube in this problem, finding the potential at the center is often straightforward because of the symmetry. The distance from the center to each charge is the same, which simplifies the summation significantly.

It's important to distinguish between electric potential (V) and electric potential energy (U). Potential is potential energy per unit charge ($V = U/q_0$). If we were to place a small test charge $q_0$ at the center of this cube, its electric potential energy would be $U = q_0 V_{total}$.

Another point to remember is that electric potential is defined relative to a reference point, usually infinity, where the potential is taken to be zero.

Prepp Team · Answer

4Q/$\sqrt{3}$​πϵ0​a

Prepp Team · Answer

Q/2$\sqrt{3}$πϵ0​a

Prepp Team · Answer

8Q/$\sqrt{3}$πϵ0​a