Animated Solution for Physics - Electrostatics: Charges are placed on the vertices of a square as shown. Let $\mathbf{E}$ be the electric field and $V$ be the potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then

A Square of Charges

Imagine a perfect square with four distinct charges pinned to its vertices: $A$, $B$, $C$, and $D$. We are tasked with understanding what happens to the electric potential $V$ and the electric field $\mathbf{E}$ at the exact geometric center of this square when we perform a specific operation: swapping the top charges with the bottom ones.

To make this abstract concept vividly clear, let's assume a standard dipole-like configuration where the top vertices $A$ and $B$ hold positive charges ($+q$), and the bottom vertices $C$ and $D$ hold negative charges ($-q$). This creates a strong downward electric field at the center, while the potential depends purely on the algebraic sum of these charges.

Let's first tackle the electric potential, $V$. The beauty of electric potential lies in its scalar nature. It doesn't care about directions or vectors; it only cares about the magnitude of the charges and their distance from the point of interest.

The formula for the potential at the center of the square is simply the sum of the potentials contributed by each individual charge:

Here, $r$ is the distance from any vertex to the center. Because it's a square, this distance $r$ is identical for all four charges. Therefore, the total potential is directly proportional to the algebraic sum of all the charges present on the square.

What Changes?

Now, the problem introduces a twist: we interchange the charges on $A$ and $B$ with those on $D$ and $C$ respectively. This means the charge that was at $A$ moves to $D$, and the charge at $D$ moves to $A$. Similarly, $B$ and $C$ swap their charges.

Let's re-evaluate our potential equation with this new arrangement:

Notice something? The terms inside the parenthesis are exactly the same, just in a different order. Addition is commutative, meaning $q_A+q_D=q_D+q_A$. Therefore, the total sum of the charges remains absolutely unchanged. Consequently, the electric potential at the center remains exactly the same: $V_{final}=V_{initial}$.

While the potential remained blissfully unaffected by the swap, the electric field $\mathbf{E}$ tells a very different story. The electric field is a vector quantity. It possesses both magnitude and direction, and it is highly sensitive to the spatial arrangement of charges.

Initially, with positive charges at the top and negative charges at the bottom, the electric field lines push away from the top and pull towards the bottom. This results in a net electric field vector pointing straight down.

When we swap the charges, we are essentially flipping the entire charge distribution upside down. Now, the positive charges are at the bottom ($D$ and $C$) and the negative charges are at the top ($A$ and $B$).

What happens to the electric field? It must now point from the new positive charges at the bottom towards the new negative charges at the top. The net electric field vector has completely reversed its direction!

In physics, if a vector changes its direction, the vector itself is considered to have changed, even if its length (magnitude) remains constant.

Therefore, our final conclusion is elegant and profound: The electric field $\mathbf{E}$ changes because its direction reverses, while the electric potential $V$ remains completely unchanged because the total scalar sum of the charges is conserved. This perfectly highlights the fundamental difference between scalar and vector fields in electrostatics.