Problem 10 - Master Resource Book in JEE Main Physics

Question

A hollow charged metal sphere has radius \(r\). If the potential difference between its surface and a point at a distance \(3 r\) from the centre is \(V\), then electric field intensity at a distance \(3 r\) is (a) \(\frac{v}{2 r}\) (b) \(\frac{v}{3 r}\) (c) \(\frac{V}{6 r}\) (d) \(\frac{V}{4 r}\)

Step-by-Step Solution

Answer

The electric field intensity at distance \(3r\) is \(\frac{V}{6r}\), option (c).

Step 1: Understanding Electric Potential

The potential difference given is between the surface of the sphere at distance \(r\) and a point at distance \(3r\) from the center. Since the sphere is a conductor, the potential on its surface is constant and can be considered as \(V_1\), and let the potential at \(3r\) be \(V_2\). The potential difference \(V = V_1 - V_2\).

Step 2: Electric Potential Relationship

The potential \(V_1\) on the surface of the sphere due to its charge \(Q\) is given by \(V_1 = \frac{kQ}{r}\), where \(k\) is Coulomb's constant. Similarly, at \(3r\), the potential \(V_2 = \frac{kQ}{3r}\). Substitute these into the potential difference equation: \(V = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r}\).

Step 3: Solving for Charge

Rearrange the derived equation for potential difference to solve for the charge \(Q\). From \(V = \frac{2kQ}{3r}\), we have \(Q = \frac{3Vr}{2k}\).

Step 4: Electric Field Calculation

The electric field \(E\) at a distance \(3r\) for a charged sphere is given by \(E = \frac{kQ}{(3r)^2}\). Substitute \(Q = \frac{3Vr}{2k}\) into this equation: \(E = \frac{k \left( \frac{3Vr}{2k} \right)}{9r^2} = \frac{3V}{18r} = \frac{V}{6r}\).

Step 5: Verify and Choose Correct Answer

The electric field intensity \(E\) at a distance \(3r\) is \(\frac{V}{6r}\), which matches option (c).

Key Concepts

Electric Potential

Electric potential is a concept that helps us understand how much energy per unit charge is stored in an electric field. Think of it like the potential energy per kilogram might have due to gravity when it is at a certain height.
In the context of the charged sphere, the electric potential at any point is related to the work done in bringing a unit positive charge from infinity to that point. It is often denoted by the symbol \( V \), and its unit is volts (V).

• The electric potential due to a point charge can be calculated using the formula: \[ V = \frac{kQ}{r} \]
• Here, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge.
• For a hollow charged sphere, this potential remains constant at the surface.

Understanding electric potential is crucial because it sets the stage for calculating the potential difference, which is essential when determining electric field intensity.

Charged Sphere

A charged sphere is a classic object of study in electrostatics, especially when looking at conductive materials. Conductors, like metal spheres, perfectly distribute any excess charge across their surface.
This homogeneous charge distribution results in several unique properties:

• Inside a conductive sphere, the electric field is zero. This is because the charges reside entirely on the surface.
• Outside the sphere, it acts like a point charge located at its center.
• The potential on the surface of the sphere is the same everywhere and can be calculated using the formula: \[ V_{surface} = \frac{kQ}{r} \]

When dealing with potential calculations, it is essential to note these properties that simplify many problems involving charged spheres.

Coulomb's Law

Coulomb's Law describes how electric charges interact. It is a foundational principle in electromagnetism, defining the force between two charges.
The law is typically expressed in the formula:

• \[ F = \frac{k |Q_1 Q_2|}{r^2} \]
• \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

For electric field calculations, Coulomb's Law helps derive the expression for the electric potential and electric field intensity:

• The electric field, which describes how a charge influences surrounding space, can be expressed as\( E = \frac{k|Q|}{r^2} \).

Understanding Coulomb's Law is vital for grasping the relationship between electric potential, field intensity, and forces exerted by charges.

Potential Difference

Potential difference, often referred to as voltage, is the difference in electric potential between two points in space. It measures how much work is done in moving a charge between these points.
This exercise uses the concept of potential difference to find the electric field intensity at a certain point. Here’s how it works:

• The potential difference \( V \) between the sphere's surface and a point \( 3r \) away is calculated as \( V = V_1 - V_2 \).
• Using the known potentials, the equation becomes \( V = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r} \).
• This helps in deducing other quantities like charge \( Q \) and electric field intensity \( E \).

By understanding potential differences, students can better understand how energy moves through fields and how potential relationships govern electric charges' influences on their surroundings.