Ten charges are placed on the circumference of a circle of radius R with constant angular separation

Q: Why is the potential zero? Don't the charges have different signs?

Exactly! Potential is a scalar, meaning it has no direction. Since all charges are at the same distance $R$, we just algebraically add the charges. Five $+q$ and five $-q$ sum to exactly zero.

Q: Don't the opposite charges cancel each other's electric field?

This is a very common trap! A positive charge pushes the field away, while a negative charge pulls it in. If they are on opposite sides of the center, their fields point in the *same* direction and actually add up, not cancel!

Q: How did we get 5 vectors spaced by $72^\circ$?

There are 10 charges, so the angle between adjacent charges is $36^\circ$. When we pair diametrically opposite charges, we get 5 pairs. The field of each pair points towards the negative charge. The negative charges are at positions 2, 4, 6, 8, 10, which are spaced by $2 \times 36^\circ = 72^\circ$.

Q: Why does the sum of 5 equal vectors spaced by $72^\circ$ equal zero?

By symmetry! If you place 5 equal forces pulling symmetrically on an object, it won't move in any direction. Mathematically, they form a closed regular pentagon when added head-to-tail.

Q: What if there were 8 charges instead of 10?

If there were 8 charges alternating $+q$ and $-q$, the angle between them would be $45^\circ$. You would get 4 pairs spaced by $90^\circ$. The vector sum would still be zero by symmetry!

Question

Ten charges are placed on the circumference of a circle of radius $R$ with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge $+q$ each, while 2, 4, 6, 8, 10 have charge $-q$ each. The potential $V$ and the electric field $E$ at the centre of the circle respectively, are (Take, $V = 0$ at infinity)

Options:

(A) $V = \frac{10q}{4\pi\varepsilon_0 R}; E = 0$

(B) $V = 0; E = \frac{10q}{4\pi\varepsilon_0 R^2}$

(C) $V = 0; E = 0$

(D) $V = \frac{10q}{4\pi\varepsilon_0 R}; E = \frac{10q}{4\pi\varepsilon_0 R^2}$

Sigma Percentile · Accepted Answer

### The Setup: A Ring of Charges
Imagine a circle with ten charges placed evenly along its boundary. We have five positive charges and five negative charges, alternating one after the other. This means the angular separation between any two adjacent charges is exactly $36^\circ$.

### Calculating the Electric Potential
Let's first find the electric potential at the center of the circle. Since electric potential is a scalar quantity, it is simply the algebraic sum of the potentials due to each individual charge.

Because all charges are at the exact same distance $R$ from the center, we can factor out the common term $\frac{k}{R}$.

$$V = \frac{k}{R} (q_1 + q_2 + \dots + q_{10})$$

Inside the bracket, we just add up all the charges. We have five $+q$ charges and five $-q$ charges.

$$V = \frac{k}{R} (5q - 5q) = 0$$

They perfectly cancel each other out! So, the net electric potential at the center is exactly zero.

### The Trap: Electric Field of Opposite Charges
Now for the electric field. Unlike potential, electric field is a vector, which means we must carefully consider directions. To make our calculation easier, let's look at pairs of charges that sit diametrically opposite to each other.

**This is where a classic mistake happens!** Many students assume that because the charges are symmetrically placed, the opposite charges will cancel each other out. But let's look closer.

Take charge 1, which is $+q$, and charge 6, which is $-q$. They sit exactly opposite to each other. The positive charge pushes the electric field *away* from it, while the negative charge pulls the electric field *towards* itself.

$$\vec{E}_{1,6} = \vec{E}_1 + \vec{E}_6$$

So, both fields point in the exact same direction! They don't cancel; they reinforce each other to give a combined vector.

$$|\vec{E}_{1,6}| = \frac{kq}{R^2} + \frac{kq}{R^2} = \frac{2kq}{R^2}$$

### The Magic of Symmetry
If we repeat this logic for all five opposite pairs, we get five identical electric field vectors. Each vector has a magnitude of $\frac{2kq}{R^2}$ and points towards the negative charge of its respective pair.

Since the ten original charges are equally spaced by $36^\circ$, these five resulting vectors are equally spaced by exactly $72^\circ$.

Now, visualize these five vectors radiating from the center. They form a perfectly symmetric star, or a regular pentagon.

**In physics, whenever you have equal vectors symmetrically distributed around a full $360^\circ$, their vector sum is always perfectly zero.**

$$\vec{E}_{net} = 0$$

So, the net electric field at the center is also zero. Both the electric potential and the electric field vanish at the center, making option (c) the correct answer. This is a beautiful problem that shows how symmetry can simplify seemingly complex vector additions!

Select Answer:

Visualized Solution

Visualizing the Setup

Electric Potential Formula

Electric Potential Calculation

Electric Field Vector Nature

Electric Field of Opposite Pairs

Electric Field of All Pairs

Net Vector Sum of Electric Field

Final Conclusion

The Sigma Insight: Electric Field

The Setup

Calculating the Electric Potential Let's first find the electric potential at the center of the circle

The Trap

The Magic of Symmetry If we repeat this logic for all five opposite pairs, we get five identical electric field vectors

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