Problem 26 - Master Resource Book in JEE Main Physics

Question

A non-conducting ring of radius \(0.5 \mathrm{~m}\) carries total charge of \(1.11 \times 10^{-10}\) C distributed non-uniformly on its circumference producting an electric field everywhere in space. The value of the line integral \(\oint_{l=\infty}^{l=0}-E \cdot d l(l=0\), being centre of ring) in volt is (a) \(+2\) (b) \(-1\) (c) \(-2\) (d) zero

Step-by-Step Solution

Answer

The correct answer is (d) zero.

Step 1: Understand the Problem

We have a non-conducting ring with a non-uniform charge distribution producing an electric field. We need to determine the value of the line integral \(\oint_{l=\infty}^{l=0}-E \cdot d l\), where \(l=0\) is the center of the ring. The line integral of the electric field represents the potential difference in electric potential between two points.

Step 2: Recall Electric Potential and Line Integral Relation

The electric potential difference (voltage) between two points is given by the integral of the electric field along the path connecting these points: \( V = \int{E \cdot dl} \). For a large distance from the ring (at infinity), the electric potential due to the ring becomes zero as electric field lines diverge.

Step 3: Evaluate Potential Difference from Infinity to the Center

The line integral from infinity to any point is equivalent to the work needed to bring a unit charge from infinitely far to that point. Since the total potential at infinity is zero, the potential at the center of the ring is equal to the negative of the potential difference from infinity to the center. Thus, we are ultimately interested in the potential solely at the center with respect to infinity.

Step 4: Deduce the Potential at the Center of the Ring

For any ring with a uniform or non-uniform charge, due to symmetry, the electric field intensity directly at its center is zero as field contributions from all parts of the ring cancel out along the plane of the ring. The potential, however, reflects cumulative work done by the field along some path.

Step 5: Analyze Impact of Non-uniform Charge on Potential

Despite the charge being non-uniform, the symmetry principle ensures that the potential at the center, along the path from infinity, only depends on the total charge and geometry (position, distance) and is additive from under an ideal ring considering similar conditions.

Step 6: Calculate or Relate Integral to Known Data

From theoretical and empirical relationships for a ring: \( V = - \frac{kQ}{R} \) at the center where \(R = 0\), \(k\) is the Coulomb's constant and considering symmetries, effects would theoretically result in zero potential change directly at the center. This conceptual deduction leads us to conclude for such symmetry \(\oint_{l=\infty}^{l=0}-E \cdot d l = 0\).

Step 7: Confirm Final Result by Logical Consistency

The choice reflects by self-corrective deductions from consistent Coulomb and potential theory concepts of symmetries, ensuring the closure of the integral over unevaluatable symmetrical axis remains zero or neutrally potential. Choose option (d) zero.

Key Concepts

Electric Field

The electric field represents the force per unit charge experienced by a positive test charge in space. It's a vector field, meaning it has both magnitude and direction. The nature of the electric field created by a non-uniform charge distribution is influenced by the charge's arrangement. In this exercise, we have a non-conducting ring with charge distributed non-uniformly around its circumference.
A non-uniform charge distribution means the electric field will vary in magnitude and direction at different points around the ring. The field is calculated using Coulomb's Law, but for unusual geometries like rings, integration is often necessary to sum up individual field vectors created by each charge element.

• The electric field at any point on the axis of the ring is derived from the vector sum of fields due to all the infinitesimal charge elements on the ring.
• At the center of the ring, however, the symmetrical arrangement can often mean that the field exactly cancels out.

Despite the complexity of calculations elsewhere, directly at the center of a ring with symmetrical distribution, even if non-uniform, the net electric field is zero.

Line Integral

A line integral in the context of electric fields involves summing up the electric field along a specific path. It's a way to calculate work done when moving a charge through an electric field. The line integral from infinity to the center of the ring measures the voltage or electric potential difference between these two points.
To put it simply, the line integral \(\oint_{l=\infty}^{l=0}-E \cdot d l\)\ considers: - The electric field strength at each infinitesimally small segment of the path - The direction of the field relative to the direction of movement

• This integral gives the work done on a unit positive charge moved from far away to the center.
• When the endpoint of the path is exactly at the center of the ring, where the net field is zero, the work done results from the total path and interaction of field flux along it.

With the potential at infinity effectively zero, the potential at the ring’s center relative to infinity is deduced, confirming with symmetry that the potential change over such an ideal path results in zero.

Non-uniform Charge Distribution

Non-uniform charge distribution means the charge is distributed unevenly over a surface or along a line. In this case, the charge on the ring is not evenly spread, possibly leading to variations in the local electric field strength as you move around the ring.
This irregular distribution affects calculations of electric potential and field greatly, as it necessitates considering a more detailed integration of components depending on the specific distribution. However, for many symmetrical problems, certain results remain straightforward through the symmetry of systems, allowing simplifications:

• The total electric charge and its effects are distributed but additively calculate to affect potential and fields as if some parts contribute less or more.
• This means considering the whole system effect rather than each segment as isolated yields a valid overview for path integrals to endpoints such as the center.

Overall, while individual calculations might seem complex due to the distribution, symmetries inherent in such systems - like in our ring - often bring us back to simplified results like zero net potential difference.