Question

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located a distance (a) \(r > R\) from the center of the shell (outside the shell); (b) \(r < R\) from the center of the shell (inside the shell).

Short answer

For \( r > R \), force is towards the center: \( F = \frac{k |qQ|}{r^2} \). For \( r < R \), the force is zero.

Step-by-step solution

Understand the Problem Statement

We need to calculate the electrostatic force on a charge \( q \) caused by a spherical shell with a uniformly distributed charge \( -Q \). The task is to consider two different positions for charge \( q \): outside the shell (when \( r > R \)) and inside the shell (when \( r < R \)).

Apply Gauss's Law for Outside the Shell \( (r > R) \)

For a charge external to a uniformly charged spherical shell, Gauss's Law states that the shell behaves as if all its charge \( -Q \) is concentrated at its center. Thus, to find the force on \( q \), consider the total charge as a point charge at the shell's center.Using Coulomb's Law:\[F = \frac{k \cdot |q| \cdot |-Q|}{r^2}\]Where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)).

Calculate the Force Magnitude for \( r > R \)

Substitute the known values into the Coulomb's Law equation:\[F = \frac{8.99 \times 10^9 \, |q| \cdot |Q|}{r^2}\]Since both charges have opposite signs the force will be attractive.

Determine the Force Direction for \( r > R \)

The direction of the force will be towards the center of the spherical shell, as the positive charge \( q \) is attracted to the negative charge \(-Q\).

Apply Gauss's Law for Inside the Shell \( (r < R) \)

According to Gauss's Law, the electric field inside a uniformly charged spherical shell (\( r < R \)) is zero. This means there is no net electrostatic force acting on a charge inside the shell.

Conclude with Force for \( r < R \)

Since the electric field inside the shell is zero, the force on the charge \( q \) is:\[F = 0\] as there is no field to exert force on the charge.

Key concepts

Coulomb's Law

Coulomb's Law is a principle that describes the electrostatic force between two charged objects. Imagine two charges, one positive and one negative, separated by a distance. Coulomb's Law helps us calculate the force of attraction or repulsion between them.

The formula for Coulomb's Law is:

• \[ F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \]

Here,

• \( F \) is the magnitude of the force.
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the magnitudes of the two charges.
• \( r \) is the distance between the charges.

Coulomb's Law tells us that as the charges increase or the distance decreases, the force grows stronger. It's an essential concept for understanding interactions in electrostatics. Whether dealing with a point charge or a more complex arrangement, keeping in mind the attributes of Coulomb's Law can help solve problems related to forces and fields in electrostatics.

Gauss's Law

Gauss's Law is a powerful tool in electrostatics. It relates the electric flux through a closed surface to the charge enclosed within. Imagine a closed 3D surface, like a bubble, surrounding some charges. Gauss's Law helps us determine the electric field resulting from those charges.

Formally, Gauss's Law is expressed as:

• \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]

Where:

• \( \Phi_E \) is the electric flux through the surface.
• \( \mathbf{E} \) is the electric field.
• \( d\mathbf{A} \) is a differential area on the closed surface.
• \( Q_{\text{enc}} \) is the total charge enclosed within the surface.
• \( \varepsilon_0 \) is the vacuum permittivity, approximately \( 8.854 \times 10^{-12} \; \text{C}^2/\text{Nm}^2 \).

Gauss's Law can greatly simplify finding electric fields. For instance, in symmetrical cases like spherical geometries, it can reduce complex calculations to manageable ones. The key takeaway is that for a charge outside a spherical shell, the shell behaves like a point charge at its center. Conversely, inside a uniformly charged shell, the electric field is zero. This understanding helps solve problems related to charged spheres and other geometries.

Spherical Shell

A spherical shell in electrostatics refers to a thin layer of charge spread evenly over the surface of a sphere. Imagine a hollow ball or bubble, where the charge distribution is only on its skin, and the interior is empty.

Understanding how a spherical shell behaves with charged particles is crucial. Especially when applying laws like Gauss's and Coulomb's. For a point charge outside of the spherical shell,

• The shell's effect is as though all its charge were concentrated at its center.
• This results in an electric field similar to a point charge, calculable through Coulomb's Law.

For a point charge located inside the shell,

• Gauss's Law reveals that the internal electric field is zero.
• The charge inside feels no net force from the shell, indicating a unique and important electrostatic property.

This behavior of spherical shells assists in simplifying electrostatic problems. By recognizing these properties, one can predict the interactions of charges with or around spherical shells with greater accuracy and ease.