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Ball in a sphere We know that if a point charge \(q\) is located at radius \(a\) in the interior of a sphere with radius \(R\) and uniform volume charge density \(\rho\), then the force on the point charge is effectively due only to the charge that is located inside radius \(a\). (a) Consider instead a uniform ball of charge located entirely inside a larger sphere of radius \(R\). Let the ball's radius be \(b\), and let its center be located at radius \(a\) in the larger sphere. Its volume charge density is such that its total charge is \(q\). Assume that the ball is superposed on top of the sphere, so that all of the sphere's charge is still present. Can the force on the ball be obtained by treating it like a point charge and considering only the charge in the larger sphere that is inside radius \(a\) ? (b) Would the force change if we instead remove the charge in the larger sphere where the ball is? So now we are looking at the force on the ball due to the sphere with a cavity carved out, which is a more realistic scenario.

Step by step solution

01

Analyze the setup

We have a smaller sphere of radius \( b \) and charge \( q \), located at a distance \( a \) from the center of a larger sphere of radius \( R \) with a uniform charge density \( \rho \). Keep in mind that the smaller sphere is superposed on top of the sphere, so all of the sphere's charge is considered to be present.

02

Consider the force on the smaller sphere

Since we consider the sphere to be located at the center of the larger one, we can say that the force on the smaller sphere due to charge inside radius \( a \) of the larger sphere is equal to \( F = \frac{1}{4\pi\epsilon_0} \times (\frac{q \times Q}{a^2}) \), where \( Q \) is the charge of the larger sphere inside radius \( a \) and \( \epsilon_0 \) is the permittivity of free space.

03

Homework Scenario

In this scenario, we are considering that the charge in the larger sphere has been removed. The charge \( Q \) which was there in the larger sphere due to the smaller sphere will not exist. So, the force acting on the smaller sphere will only be due to the remaining charge in the larger sphere. If the force reduces, it shows that the removal of the charge does affect the force on the smaller sphere.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field

The concept of an electric field is fundamental in understanding electrostatics. An electric field is a region around a charged object where another charge experiences a force. It is represented by field lines and is defined as the force per unit charge. The strength and direction of an electric field depend on the source charge's magnitude and sign.

Calculating the electric field at a point involves:

• Identifying the source of the electric field, which could be a point charge, line, or surface charge.
• Using the formula for electric field: \( E = \frac{F}{q} = \frac{k \, |Q|}{r^2} \), where \( Q \) is the source charge, \( r \) is the distance from the charge, and \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
• Considering the direction. The electric field vector points away from positive charges and towards negative charges.

In exercises like the one discussed, understanding where the electric field is strongest or weakest helps us predict the forces acting on the charges involved.

Gauss's Law

Gauss's Law provides a powerful tool for analyzing electric fields, particularly in symmetric situations like spheres and cylinders. The law states that the total electric flux through a closed surface equals the charge enclosed divided by the permittivity of space \( (\epsilon_0) \).

Mathematically, it is expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]

In our problem:

• The smaller sphere (ball) can be analyzed using Gauss's Law to determine the electric field within and outside the larger sphere.
• The symmetry of the problem simplifies the calculation of the electric field, as only the charge inside the Gaussian surface at radius \( a \) contributes to the field inside.
• Understanding how the charge distribution affects the electric field is crucial when considering modifications like introducing a cavity or moving the charge.

Gauss's Law helps simplify complex calculations, making it easier to see the impact of different geometries and scenarios on electric fields.

Coulomb's Law

Coulomb's Law is essential for calculating the force between two point charges. It gives the magnitude of the force as directly proportional to the product of the absolute charges and inversely proportional to the square of the distance between them.

The formula is: \[ F = \frac{1}{4\pi\epsilon_0} \times \frac{|q_1 \times q_2|}{r^2} \]
Where:

• \( q_1 \) and \( q_2 \) are the point charges involved.
• \( r \) is the distance between the charges.
• \( \epsilon_0 \) is the permittivity of free space.

For the exercise provided:

• Coulomb's Law helps us determine the force on the smaller sphere due to the portion of the larger sphere's charge that lies within a specific radius.
• The problem addresses whether this force is altered when part of the charge distribution is removed.
• Coulomb's Law underscores the importance of understanding how distance and charge magnitude affect electrostatic forces.

This principle is crucial in predicting how objects interact under the influence of electric fields created by different configurations of charges.