Six point charges, each of the same magnitude q, are arranged in different manners as shown in Colum

Q: Why is the electric field zero for the hexagon in configuration p?

Although opposite charges don't cancel each other (they actually add up), you get three net field vectors separated by $120^\circ$. The vector sum of three equal vectors at $120^\circ$ is zero.

Q: Why is the potential non-zero for configuration r?

Even though there are three positive and three negative charges, they are at different distances from the center $M$. Potential depends on distance, so they don't cancel out.

Q: How do we determine if the magnetic field is zero during rotation?

Identify the circular path each charge takes around the axis of rotation. If a positive and a negative charge share the exact same circular path, their equivalent currents cancel out.

Q: Why is the magnetic field zero for configuration t?

When rotating about the vertical axis $PQ$, every charge on the left ring has a counterpart on the right ring with the opposite sign that traces the exact same circle in space. Their currents perfectly cancel.

Q: Does a zero magnetic field always imply a zero magnetic moment?

In these highly symmetric configurations where currents perfectly cancel in every circular loop, both the magnetic field at the center and the overall magnetic moment of the system will be zero.

Question

Six point charges, each of the same magnitude $q$, are arranged in different manners as shown in Column II. In each case, a point $M$ and a line $PQ$ passing through $M$ are shown. Let $E$ be the electric field and $V$ be the electric potential at $M$ (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line $PQ$. Let $B$ be the magnetic field at $M$ and $\mu$ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

List-I:

P. $E = 0$

Q. $V 
eq 0$

R. $B = 0$

S. $\mu 
eq 0$

List-II:

1. Charges are at the corners of a regular hexagon. $M$ is at the centre of the hexagon. $PQ$ is perpendicular to the plane of the hexagon.

2. Charges are on a line perpendicular to $PQ$ at equal intervals. $M$ is the mid-point between the two innermost charges.

3. Charges are placed on two coplanar insulating rings at equal intervals. $M$ is the common centre of the rings. $PQ$ is perpendicular to the plane of the rings.

4. Charges are placed at the corners of a rectangle of sides $a$ and $2a$ and at the mid points of the longer sides. $M$ is at the centre of the rectangle. $PQ$ is parallel to the longer sides.

5. Charges are placed on two coplanar, identical insulating rings at equal intervals. $M$ is the mid points between the centres of the rings. $PQ$ is perpendicular to the line joining the centres and coplanar to the rings.

Sigma Percentile · Accepted Answer

### The Symphony of Rotating Charges: A Masterclass in Symmetry

This problem is a beautiful exploration of how spatial symmetry dictates the behavior of electric and magnetic fields. By analyzing five distinct configurations of six point charges, we can deeply understand the interplay between electrostatics and magnetism when systems are set into rotation.

### Configuration (p): The Hexagon
Imagine a regular hexagon with alternating positive and negative charges. Because there are exactly three $+q$ and three $-q$ charges, the net charge is zero, making the electric potential $V$ at the center zero.

What about the electric field? The fields from opposite charges don't cancel; they actually add up! However, we have three such pairs oriented at $120^\circ$ to each other. The vector sum of three equal vectors separated by $120^\circ$ is exactly zero. Thus, $E = 0$.

When we rotate this hexagon about the axis $PQ$, the positive and negative charges move in the exact same circular path. A positive charge moving in a circle creates a current, and a negative charge moving in the same circle creates an equal current in the opposite direction. The net current is zero, meaning both the magnetic field $B$ and magnetic moment $\mu$ are zero.

### Configuration (q): The Linear Array
Here, we have six charges on a straight line, alternating in sign. Again, the net charge is zero, so $V = 0$.

However, the electric field at the midpoint $M$ is not zero. The fields from the symmetric pairs do not cancel out perfectly because they are at different distances from $M$.

When we rotate this line about the perpendicular bisector $PQ$, symmetric pairs of $+q$ and $-q$ move in the same circular orbits. Just like the hexagon, their currents cancel each other out perfectly. Thus, $B = 0$ and $\mu = 0$.

### Configuration (r): Concentric Rings
We have two concentric rings: the inner ring has three $-q$ charges, and the outer ring has three $+q$ charges. The potential $V$ at the center is not zero because the positive and negative charges are at different distances from $M$.

But the electric field $E$ is zero! Why? Because three equal charges placed symmetrically on a ring produce zero field at the center.

When we rotate this system about $PQ$, the inner and outer rings create currents at different radii. These currents do not cancel out, so both $B 
eq 0$ and $\mu 
eq 0$.

### Configuration (s): The Rectangle
We have a rectangle with charges at the corners and midpoints. The sum of all charges is $-2q$, so the potential $V$ is definitely not zero.

But look at the symmetry! Opposite pairs relative to the center $M$ have the exact same sign and distance. Their electric fields cancel out perfectly, making $E = 0$.

When rotating about the horizontal axis $PQ$, the charges move in circular paths. The net current in these paths is non-zero, leading to $B 
eq 0$ and $\mu 
eq 0$.

### Configuration (t): Twin Rings
Finally, we have two identical rings side by side. The total charge is zero, so $V = 0$. The electric field $E$ is non-zero because the left and right sides have different net charges.

When rotating about the vertical axis $PQ$, a fascinating thing happens. For every charge on the left ring, there is a corresponding charge on the right ring with the opposite sign that rotates in the exact same circular path! Their currents cancel out completely. Thus, $B = 0$ and $\mu = 0$.

List-I

List-II

Select Matching Pairs:

Visualized Solution

$Analyzing Configuration (p)$

$Analyzing Configuration (q)$

$Analyzing Configuration (r)$

$Analyzing Configuration (s)$

$Analyzing Configuration (t)$

$Final Matching$

The Sigma Insight: Magnetic Moment of Current Loop

The Symphony of Rotating Charges

Configuration (p)

Configuration (q)

Configuration (r)

Configuration (s)

Configuration (t)

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