Animated Solution for Physics - Electrostatics: A cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin, where the charge is $-Q$. The electric field at the centre of cube is

Imagine standing in a room where every single corner is pulling you with an equal force. What happens? You don't move. You are in perfect equilibrium. This is the profound power of symmetry in physics, and it is exactly the secret weapon we will use to dismantle this seemingly complex JEE problem.

At first glance, calculating the electric field at the center of a cube with charges scattered across its vertices looks like a mathematical nightmare. We have a cube of side $a$. Seven of its corners host a positive charge $+Q$, while the origin stubbornly holds a negative charge $-Q$. If we were to calculate the electric field vector for each of these eight charges individually and sum them up, we would be drowning in a sea of components, angles, and square roots.

But as elite problem solvers, we don't work harder; we work smarter.

Superposition

The principle of superposition tells us that the net electric field is simply the vector sum of individual fields. But it also allows us to play a clever mathematical trick. What if we could artificially create perfect symmetry?

Let's focus on that odd $-Q$ charge at the origin. Mathematically, we can rewrite $-Q$ as the sum of $+Q$ and $-2Q$.

Why on earth would we do this? Because by placing a $+Q$ and a $-2Q$ at the origin, we haven't changed the physical reality of the problem (the net charge there is still $-Q$), but we have completely transformed the geometry!

Now, look at the cube. We have a $+Q$ charge at every single one of the eight corners.

Because the cube is perfectly symmetrical, the electric field produced by a $+Q$ charge at any corner is exactly canceled by the electric field produced by the $+Q$ charge at the diametrically opposite corner.

At the exact center of the cube, the vector sum of the electric fields from all eight $+Q$ charges is a beautiful, resounding zero.

Suddenly, our nightmare problem has vanished. The only charge that actually contributes to the net electric field at the center is the "leftover" $-2Q$ charge sitting at the origin.

Now, we just need to find the electric field produced by a single point charge $-2Q$ at the center of the cube. We will use the vector form of Coulomb's law:

First, we need the position vector $\mathbf{r}$ of the center of the cube relative to the origin. Since the center lies exactly halfway along each edge, its coordinates are $(a/2,a/2,a/2)$.

Next, we calculate the magnitude of this position vector, which is the distance from the origin to the center. Using the 3D distance formula:

With our components ready, we substitute $q=-2Q$, $\mathbf{r}$, and $r$ into our electric field equation:

Let's carefully expand the denominator:

Substituting this back in:

Finally, simplifying the constants by canceling the $4$ in the denominator with the $8$ in the numerator, we arrive at our elegant final answer:

The negative sign perfectly aligns with our physical intuition: since the effective charge at the origin is negative ($-2Q$), the electric field at the center must point towards the origin, which is exactly the opposite direction of the position vector $(\hat{x}+\hat{y}+\hat{z})$.

By leveraging symmetry, we bypassed pages of tedious vector addition and arrived at the solution with clarity and precision. This is the hallmark of true mastery in physics!