Problem 132 - A Complete Resource Book in Physics for JEE Main

Question

A position charge \(Q\) is uniformly distributed over the ring of radius \(R=1 \mathrm{~m}\), and potential at infinity is assumed to be zero. Column-I (A) Electric field at centre of ring is (B) Electric field at the axis of ring at a distance \(x=\sqrt{3} \mathrm{~m}\) from centre is (C) Electric potential at centre of ring is (D) Electric potential on axis of ring at a distance \(\sqrt{3} \mathrm{~m}\) from centre of ring is Column-II 1\. \(\frac{K Q}{2}\) 2\. \(K Q\) 3\. \(\frac{K Q \sqrt{3}}{8}\) 4\. Zero 5\. \(\frac{K Q \sqrt{2}}{7}\)

Step-by-Step Solution

Answer

(A) Electric field at the center of the ring: 4. Zero (B) Electric field at the axis of the ring at a distance x = √3 m: 3. \(\frac{kQ\sqrt{3}}{8}\) (C) Electric potential at the center of the ring: 2. \(kQ\) (D) Electric potential on the axis of the ring at a distance x = √3 m : 1. \(\frac{kQ}{2}\)

Step 1: Setup the variables and the necessary formulas

To solve this problem, we have to use the following variables and formulas: 1. Charge (Q) 2. Ring radius (R) 3. Electric field (E) 4. Electric potential (V) 5. Coulomb's constant (K) The formula for the electric field and potential due to a point charge is given by: \(E = \frac{kQ}{r^2}\) and \(V = \frac{kQ}{r}\) where r is the distance from the point charge. Now we will evaluate the electric field and potential at the given positions.

Step 2: Electric field at the center of the ring

At the center of the ring, all elements of charge on the ring have equal and opposite contributions to the electric field. So, when we add up all these electric fields, they will cancel each other out, resulting in a net electric field of zero.

Step 3: Electric field at the axis of the ring at a distance x=√3 m

For this case, let's consider an infinitesimal charge \(dQ\) on the ring, the electric field due to this charge at a distance x from the center is \(dE = \frac{k dQ}{r^2}\), where r is the distance from the charge to the point on the axis. The distance r can be calculated using the Pythagorean theorem: \(r^2 = x^2 + R^2\), in this case with x = √3 m, and R = 1 m, we have \(r^2 = (3 + 1) = 4\). Hence, \(dE = \frac{k dQ}{4}\). By symmetry, only the components of the electric field along the axis will add up, and the perpendicular components will cancel each other. So, the net electric field is: \(E = \int \frac{k dQ}{4} \cos{\theta}\), where θ is the angle between r and the axis. We can relate θ with the ring radius R and distance r with \(\cos{\theta} = \frac{x}{r}\). In this case, \(\cos{\theta} = \frac{\sqrt{3}}{2}\). Now, we just need to integrate the electric field due to all the charges on the ring: \(E = \int \frac{k dQ}{4} \frac{\sqrt{3}}{2}\) which can be simplified as: \(E = \frac{k Q \sqrt{3}}{8}\)

Step 4: Electric potential at the center of the ring

For the electric potential at the center of the ring, we don't have to worry about the direction of the electric field, so we can find the electric potential directly using the formula for potential due to point charge: Since all charges on the ring are equidistant from the center, the electric potential at the center will be the same for all charges and can be calculated as: \(V = \frac{kQ}{R}\) In this case, with R = 1 m, we have: \(V = kQ\)

Step 5: Electric potential at the axis at a distance x=√3 m

Similarly, we can find the electric potential at a distance x = √3 m from the center on the axis. Using the distance r calculated earlier, r = 2 m, we have: \(V = \frac{kQ}{r}\) So, for this case, \(V = \frac{kQ}{2}\) Now, we can match the given options in Column-II with the results from the calculated values in Column-I. Results: (A) Electric field at the center of the ring: 4. Zero (B) Electric field at the axis of the ring at a distance x = √3 m: 3. \(\frac{kQ\sqrt{3}}{8}\) (C) Electric potential at the center of the ring: 2. \(kQ\) (D) Electric potential on the axis of the ring at a distance x = √3 m : 1. \(\frac{kQ}{2}\)

Key Concepts

Electric Field

An electric field is a region around a charged object where other charges experience a force. It's a vector field and has both a magnitude and a direction.
The electric field (E) due to a point charge is calculated using Coulomb's Law in the form:\[E = \frac{kQ}{r^2}\]Here, E is the electric field, Q is the charge, K is Coulomb's constant (9 \, \times \, 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2), and r is the distance from the charge.

• When multiple charges exist, their electric fields combine.
• Electric fields point away from positive charges and towards negative ones.
• At the center of a symmetrically charged ring, electric fields due to individual charges cancel out, as seen in step 2 of the solution where the net electric field is zero at the center.

This is because each charge pushes in opposition with its symmetrical counterpart.

Electric Potential

Electric potential quantifies the potential energy per unit charge at a point in space due to electric fields. It's a scalar quantity and doesn't have direction.
The formula for electric potential (V) due to a point charge is:\[V = \frac{kQ}{r}\]

• Considered at points infinitely far away, the electric potential is typically assumed to be zero.
• For a ring, all charge elements at any point contribute equally to the potential, making the calculation straightforward.
• At the center of a ring with charge, potential sums up uniformly from its elements.

As mentioned in step 4, the potential at the center can then be calculated directly resulting in: \( V = kQ \).

Point Charge

A point charge is an idealized model of a charged object, where the size of the object is considered to be negligible compared to other relevant dimensions. This simplification allows calculations of electric fields and potentials easier.

• This concept helps in analyzing simpler systems and always uses the center of charge for calculations.
• It simplifies the computation of forces and potentials in electric fields.
• In most questions, especially with spherical symmetry like a charged ring, point charge calculations help deduce the electric properties even when distribution occurs.

In this problem, as in many others, each infinitesimally small part of a ring's charge distribution can be treated as a point charge to calculate potential and field individually.

Coulomb's Law

Coulomb's Law is central to understanding the force between two point charges. It states that the force (F) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them:\[F = k \frac{|Q_1 Q_2|}{r^2}\]

• K is the Coulomb's constant.
• The force is attractive if the charges are opposite, and repulsive if they are the same.
• This law applies in vacuum and is foundational for electric force calculation.

In the context of calculations in the provided exercise, even when visualizing charge distribution on a ring, the elemental forces or potentials from infinitesimal charge elements are summed up using integrals for accurate results. Coulomb's Law imparts a clearer view on these elemental force interactions.