Six point charges are kept at the vertices of a regular hexagon of side L and centre O as shown in t

Q: Why is the resultant of the 2K fields exactly 2K?

Because the angle between them is $120^\circ$. The resultant of two equal vectors $V$ at $120^\circ$ is $2V \cos(120^\circ/2) = 2V \cos(60^\circ) = V$.

Q: Why do we group the charges in pairs?

Grouping opposite charges lying on the same diagonal simplifies the vector addition, as their electric fields at the center point in the same direction.

Q: Why is the potential zero on the line PR?

The line PR acts as a mirror of symmetry. Every positive charge on the left has a corresponding negative charge of equal magnitude at the exact same distance on the right. Their scalar potentials cancel out perfectly.

Q: Can the potential be zero if the electric field is non-zero?

Absolutely! The center O is a perfect example. The potential is zero because the scalar sum of charges is zero, but the electric field is non-zero because the vector sum of the fields is not zero.

Q: Why does the potential approach infinity near the charges?

Electric potential is inversely proportional to the distance from the charge ($V = kq/r$). As $r$ approaches zero, the potential approaches infinity.

Question

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ as shown in the figure. Given that $K = \frac{1}{4\pi\varepsilon_0} \frac{q}{L^2}$, which of the following statements(s) is(are) correct.

Options:

(A) The electric field at $O$ is $6K$ along $OD$

(B) The potential at $O$ is zero

(C) The potential at all points on the line $PR$ is same

(D) The potential at all points on the line $ST$ is same

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the center of a regular hexagon, surrounded by six point charges. This is a classic electrostatics problem that tests your understanding of both vector fields and scalar potentials.

The charges are arranged symmetrically, but their magnitudes and signs vary. We have $+2q$ at $A$, $+q$ at $B$, $-q$ at $C$, $-2q$ at $D$, $-q$ at $E$, and $+q$ at $F$.

Our goal is to evaluate the electric field and potential at the center $O$, as well as the potential along two specific lines, $PR$ and $ST$. Let's break this down step by step.

### The Master Equation for Electric Field

Let's start by calculating the electric field at the center $O$. The electric field is a vector, so direction matters immensely. We can simplify our work by pairing up charges that lie on the same diagonal.

First, consider the charges at $A$ and $D$. The positive charge $+2q$ at $A$ pushes a positive test charge at $O$ towards $D$ with a field magnitude of $2K$. Simultaneously, the negative charge $-2q$ at $D$ pulls the test charge towards itself with another $2K$.

$$E_{AD} = 2K + 2K = 4K$$

This combined field of $4K$ points directly towards $D$.

### Evaluating the Other Diagonals

Next, let's look at the diagonal connecting $F$ and $C$. The charge $+q$ at $F$ pushes towards $C$ with a field $K$, and the charge $-q$ at $C$ pulls towards $C$ with a field $K$.

$$E_{FC} = K + K = 2K$$

This field points towards $C$. Similarly, for the diagonal connecting $B$ and $E$, the charge $+q$ at $B$ pushes towards $E$, and $-q$ at $E$ pulls towards $E$.

$$E_{BE} = K + K = 2K$$

This field points towards $E$. Now we have three vectors at the center: $4K$ towards $D$, $2K$ towards $C$, and $2K$ towards $E$.

### The Final Vector Addition

The angle between the vectors pointing towards $C$ and $E$ is exactly $120^\circ$. The resultant of two equal vectors of magnitude $2K$ separated by $120^\circ$ is simply $2K$, and it bisects the angle, pointing straight towards $D$.

$$E_{resultant} = \sqrt{(2K)^2 + (2K)^2 + 2(2K)(2K)\cos(120^\circ)} = 2K$$

Adding this to our initial $4K$ field (which also points towards $D$), we get the total electric field at $O$.

$$E_{net} = 4K + 2K = 6K$$

The net electric field is **$6K$ along $OD$**. Thus, statement (a) is absolutely correct.

### The Elegance of Scalar Potential

Now, let's shift our focus to the electric potential at the center $O$. Unlike the electric field, potential is a scalar quantity. We don't need to worry about directions; we just add the values algebraically.

Since all the vertices of a regular hexagon are at the same distance $L$ from the center, the potential is simply proportional to the algebraic sum of all the charges.

$$V_O = \frac{k}{L} \sum q_i$$

Let's sum the charges: $+2q + q - q - 2q - q + q$.

$$V_O = \frac{k}{L} (0) = 0$$

The total charge is zero, so the potential at the center is exactly **zero**. Statement (b) is correct.

### Symmetry on the Line PR

What about the potential on the line $PR$? This line connects the midpoints of the top and bottom sides of the hexagon. Geometrically, it acts as the perpendicular bisector for the horizontal segments $AD$, $FE$, and $BC$.

For any point on this line, the distance to a charge on the left is perfectly equal to the distance to the corresponding charge on the right.

$$V_{PR} = k \left( \frac{2q - 2q}{r_1} + \frac{q - q}{r_2} + \frac{q - q}{r_3} \right) = 0$$

Because the left half of the hexagon is a positive mirror image of the negative right half, their potentials cancel out everywhere along $PR$. The potential is **zero at all points on $PR$**. Statement (c) is correct.

### The Asymmetry of Line ST

Finally, let's examine the line $ST$, which lies along the horizontal axis passing through $A$ and $D$.

As you move along this line towards $A$, you get infinitely close to the $+2q$ charge, causing the potential to approach $+\infty$. Conversely, as you move towards $D$, you approach the $-2q$ charge, and the potential drops to $-\infty$.

Clearly, the potential is **not constant** along the line $ST$. Therefore, statement (d) is incorrect.

In conclusion, the correct statements are **(a), (b), and (c)**. This problem is a beautiful reminder of how symmetry can drastically simplify complex physical systems!

Animated Solution for Physics - Electrostatics: Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$ as shown in the figure. Given that $K = \frac{1}{4 π ε _{0}} \frac{q}{L ^{2}}$ , which of the following statements(s) is(are) correct.

Select Answer:

Visualized Solution

$E$ -field due to $A$ and $D$

$E$ -field due to other pairs

Resultant $E$ -field at $O$

Potential at $O$

Potential on line $P R$

Potential on line $S T$

Conclusion

The Sigma Insight: Electric Field

Analyzing the Setup

The Master Equation for Electric Field

Evaluating the Other Diagonals

The Final Vector Addition

The Elegance of Scalar Potential

Symmetry on the Line PR

The Asymmetry of Line ST

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