Problem 51 - Arihant AIEEE Physics

Question

A non-conducting solid sphere of radius \(R\) is uniformly charged. The magnitude of electric field due to the sphere at a distance \(r\) from its centre: (a) increases as \(r\) increases for \(r

Step-by-Step Solution

Answer

(e) both (a) and (c) are correct.

Step 1: Understanding the Problem

We need to analyze how the electric field behaves at different distances from the center of a uniformly charged non-conducting solid sphere. We need to identify which options correctly describe the behavior of the electric field at those distances.

Step 2: Electric Field Inside the Sphere

For points inside the sphere, where \(r < R\), the electric field \(E\) is given by the formula \(E = \frac{kQr}{R^3}\), where \(k\) is Coulomb’s constant and \(Q\) is the total charge. The electric field increases linearly with \(r\) as you move away from the center towards the surface of the sphere.

Step 3: Electric Field Outside the Sphere

For points outside the sphere, where \(r > R\), the electric field is given by the expression \(E = \frac{kQ}{r^2}\). In this region, the electric field decreases with increasing \(r\), following an inverse square law, similar to a point charge.

Step 4: Evaluate the Options

Now, let's match the behavior we've deduced with the provided options: - (a) is correct because the electric field increases with \(r\) for \(r

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that governs the interaction between electrically charged particles. The law states that the electric force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

This can be mathematically represented as:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:

• \( F \) represents the magnitude of the electric force
• \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \) Nm²/C²
• \( q_1 \) and \( q_2 \) are the amounts of the two charges
• \( r \) is the distance between the centers of the two charges

Coulomb's Law helps explain why charged particles either attract or repel each other, depending on their types of charges. This principle is crucial in understanding how the electric field behaves around charged objects such as spheres.

Electric Field Inside a Sphere

The electric field inside a uniformly charged non-conducting solid sphere behaves differently than one might expect from a point charge. For any point within the sphere, the field strength can be found using Gauss's Law, but an intuitive understanding is also helpful.

Inside the sphere, at a distance \( r \) from the center, the electric field \( E \) is calculated by:
\[ E = \frac{k Q r}{R^3} \]
where:

• \( E \) is the electric field at distance \( r \)
• \( Q \) is the total charge of the sphere
• \( R \) is the radius of the sphere
• \( k \) is Coulomb's constant

As you move further from the center towards the outer surface within the sphere \((r

Electric Field Outside a Sphere

For points outside a non-conducting charged sphere \((r > R)\), the electric field behaves similarly to that around a point charge. This is because, according to Gauss's Law, when considering an electric field outside a charged sphere, the entire charge of the sphere can be considered as concentrated at its center.

The formula for the electric field at a distance \( r \) from the center of the sphere is:
\[ E = \frac{k Q}{r^2} \]

In this scenario:

• \( E \) represents the electric field at distance \( r \)
• \( Q \) is the total charge on the sphere
• \( k \) is Coulomb's constant

The electric field decreases with the square of the distance \((r)\). This means as you move further away, the influence of the sphere's charge weakens rapidly. This inverse square law behavior is characteristic of the electric field due to point charges and spherical charge distributions like our sphere here. Thus, the electric field outside of a sphere decreases as \( r \) increases, highlighting the diminishing effect of the charge with distance.

Previous

Problem 50

Next

Problem 52

Other exercises in this chapter

Problem 49

Three charged particles are collinear and are in equilibrium. Then: (a) all the charged particles have the same polarity (b) the equilibrium is unstable(c) all

View solution

Problem 50

A point charge \(q\) and a charge \((-q)\) are placed at \(x=-a\) and \(x=+a\) respectively. Which of the following represents a part of \(E-x\) graph ?

View solution

Problem 52

A conducting solid sphere of radius \(R\) is uniformly charged. The magnitude of the electric field due to the sphere at a distance \(r\) from its centre (a) in

View solution

Problem 57

Calculate the work done in carrying a charge \(q\) once round over a closed circular path of radius ' \(r^{\prime}\) and a charge \(Q\) is at the centre : (a) \

View solution

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that governs the interaction between electrically charged particles. The law states that the electric force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

This can be mathematically represented as:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:

• \( F \) represents the magnitude of the electric force
• \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \) Nm²/C²
• \( q_1 \) and \( q_2 \) are the amounts of the two charges
• \( r \) is the distance between the centers of the two charges

Coulomb's Law helps explain why charged particles either attract or repel each other, depending on their types of charges. This principle is crucial in understanding how the electric field behaves around charged objects such as spheres.

Electric Field Inside a Sphere

The electric field inside a uniformly charged non-conducting solid sphere behaves differently than one might expect from a point charge. For any point within the sphere, the field strength can be found using Gauss's Law, but an intuitive understanding is also helpful.

Inside the sphere, at a distance \( r \) from the center, the electric field \( E \) is calculated by:
\[ E = \frac{k Q r}{R^3} \]
where:

• \( E \) is the electric field at distance \( r \)
• \( Q \) is the total charge of the sphere
• \( R \) is the radius of the sphere
• \( k \) is Coulomb's constant

As you move further from the center towards the outer surface within the sphere \((r

Electric Field Outside a Sphere

For points outside a non-conducting charged sphere \((r > R)\), the electric field behaves similarly to that around a point charge. This is because, according to Gauss's Law, when considering an electric field outside a charged sphere, the entire charge of the sphere can be considered as concentrated at its center.

The formula for the electric field at a distance \( r \) from the center of the sphere is:
\[ E = \frac{k Q}{r^2} \]

In this scenario:

• \( E \) represents the electric field at distance \( r \)
• \( Q \) is the total charge on the sphere
• \( k \) is Coulomb's constant

The electric field decreases with the square of the distance \((r)\). This means as you move further away, the influence of the sphere's charge weakens rapidly. This inverse square law behavior is characteristic of the electric field due to point charges and spherical charge distributions like our sphere here. Thus, the electric field outside of a sphere decreases as \( r \) increases, highlighting the diminishing effect of the charge with distance.

Previous

Problem 50

Next

Problem 52

Other exercises in this chapter

Problem 49

Three charged particles are collinear and are in equilibrium. Then: (a) all the charged particles have the same polarity (b) the equilibrium is unstable(c) all

View solution

Problem 50

A point charge \(q\) and a charge \((-q)\) are placed at \(x=-a\) and \(x=+a\) respectively. Which of the following represents a part of \(E-x\) graph ?

View solution

Problem 52

A conducting solid sphere of radius \(R\) is uniformly charged. The magnitude of the electric field due to the sphere at a distance \(r\) from its centre (a) in

View solution

Problem 57

Calculate the work done in carrying a charge \(q\) once round over a closed circular path of radius ' \(r^{\prime}\) and a charge \(Q\) is at the centre : (a) \

View solution