Problem 101 - Master Resource Book in JEE Main Physics

Question

A regular hexagon of side \(10 \mathrm{~cm}\) has a charge \(5 \mu \mathrm{C}\) at each of its vartices. The potential at the centre of the hexagon is? [NCERT] (a) \(3.7 \times 10^{6} \mathrm{~V}\) (b) \(2.7 \times 10^{6} \mathrm{~V}\) (c) \(4 \times 10^{6} \mathrm{~V}\) (d) \(5 \times 10^{7} \mathrm{~V}\)

Step-by-Step Solution

Answer

The potential at the center is \(2.7 \times 10^{6} \mathrm{~V}\) (option b).

Step 1: Identify Total Charge

Since there are 6 vertices and each vertex has a charge of \(5 \mu \mathrm{C}\), the total charge at the vertices is the sum of all charges: \[ Q_{total} = 6 \times 5 \mu \mathrm{C} = 30 \mu \mathrm{C}. \] However, for potential calculation, it's more useful to consider each vertex separately.

Step 2: Calculate Distance from Center to Vertex

The distance from the center of a regular hexagon to one of its vertices is the same as the side of the hexagon, which is \(10 \mathrm{~cm} = 0.1 \mathrm{~m}\).

Step 3: Calculate Potential Due to One Charge at the Center

The potential \(V\) at the center due to one charge \(q\) is given by the formula:\[ V = \frac{kq}{r}, \]where:- \(k = 9 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2\) is Coulomb's constant,- \(q = 5 \mu \mathrm{C} = 5 \times 10^{-6} \mathrm{~C}\),- \(r = 0.1 \mathrm{~m} \) is the distance from the charge to the center.Thus, the potential due to one charge is:\[ V_{1} = \frac{(9 \times 10^9) (5 \times 10^{-6})}{0.1} = 4.5 \times 10^5 \mathrm{~V}. \]

Step 4: Calculate Total Potential at the Center

Since the total potential at a point due to multiple charges is the algebraic sum of the potentials due to each charge, and there are 6 equal charges positioned symmetrically:\[ V_{total} = 6 \times V_{1} = 6 \times 4.5 \times 10^5 = 2.7 \times 10^6 \mathrm{~V}. \]

Key Concepts

Regular Hexagon

A regular hexagon is a six-sided polygon where all sides are of equal length and all interior angles are the same. Hexagons often appear in nature, such as in honeycombs, or in engineering and architecture, due to their efficient geometric properties.

The study of regular hexagons is crucial when dealing with problems involving symmetry, especially in physics. In our exercise, each vertex of the hexagon holds a charge. A regular hexagon allows for simple calculations of symmetrical electric field and potential since each side and vertex is uniquely similar and predictable relative to the center.

The trick to tackling geometric problems such as this one is recognizing the repeating patterns that make the math elegant and manageable.

Coulomb's Law

Coulomb's law is a foundational principle in electrostatics. It quantifies the amount of force between two stationary, electrically charged particles.

According to Coulomb's law, the electrostatic force (F) between two point charges is directly proportional to the product of the absolute magnitudes of each charge, and inversely proportional to the square of the distance between the charges. The formula is given as:

• \( F = \frac{k |q_1 q_2|}{r^2} \)

Where:

• \( k \) is the electrostatic constant, approximately \( 9 \times 10^9 \text{ N m}^2/ ext{C}^2 \).
• \( q_1 \) and \( q_2 \) are the values of each charge.
• \( r \) is the distance between the charges.

In the context of the exercise, this law helps calculate the electric potential due to one charge. However, the potential at the center is the sum of potentials due to all six charges, exploiting the hexagon's symmetry.

Symmetry in Physics

Symmetry plays an important role in simplifying complex physical systems. When charges are arranged symmetrically, like in a regular hexagon, each charge's contribution to the potential at a particular point can be perceived in a balanced way.

By using symmetry, we can reduce the complexity involved in calculations by identifying repetitive or reflective patterns. In this exercise, each charge contributes equally to the potential at the hexagon's center.

Because of the consistent arrangement in a hexagon, calculations use symmetry to simplify the addition of potentials. This makes working with the charges straightforward, as the uniformity provides a configuration where each component behaves predictably in relation to others. Hence, understanding symmetry allows students to quickly grasp why some physical systems behave the way they do.