Problem 40 - University Physics with Modern Physics

Question

22.40. A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\) (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho .\) (b) What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R\) (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R\) .

Step-by-Step Solution

Answer

(a) \( E = \frac{\rho r}{2\varepsilon_0} \); (b) \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \); (c) Electric fields are equal at \( r=R \); (d) Field rises linearly inside, decreases inversely outside.

Step 1: Determine Electric Field Inside the Cylinder

To find the electric field inside the cylinder at a distance \( r \) from the axis, we use Gauss's Law. Consider a cylindrical Gaussian surface of radius \( r \) and length \( L \) coaxial with the charged cylinder. The electric field \( E \) is radial, and its magnitude is constant on the Gaussian surface: \[\Phi_E = E(2\pi r L) = \frac{Q_{enc}}{\varepsilon_0}\]The charge enclosed \( Q_{enc} \) by the Gaussian surface is given by the charge density \( \rho \) times the volume of the cylinder: \[Q_{enc} = \rho \pi r^2 L\]Substituting \( Q_{enc} \) in the Gauss's law equation:\[E(2\pi r L) = \frac{\rho \pi r^2 L}{\varepsilon_0}\]Solving for \( E \):\[E = \frac{\rho r}{2\varepsilon_0}\]

Step 2: Determine Electric Field Outside the Cylinder

For the electric field outside at a distance greater than \( R \), we again employ Gauss's Law. This time, consider a cylindrical Gaussian surface of radius \( r \) and length \( L \) (\( r > R \)):\[\Phi_E = E(2\pi r L) = \frac{Q_{enc}}{\varepsilon_0}\]The total charge \( Q_{enc} \) in the original charged cylinder is given by: \[Q = \rho \pi R^2 L\]Or, it can be expressed in terms of charge per unit length \( \lambda \):\[\lambda = \rho \pi R^2\]Thus,\[E(2\pi r L) = \frac{\lambda L}{\varepsilon_0}\]Solving for \( E \):\[E = \frac{\lambda}{2\pi \varepsilon_0 r}\]

Step 3: Compare Electric Fields for \( r = R \)

Substitute \( r = R \) in the expressions for electric fields derived in Steps 1 and 2:- Inside (from part a): \[E_{inside} = \frac{\rho R}{2\varepsilon_0}\]- Outside (from part b): \[E_{outside} = \frac{\lambda}{2\pi \varepsilon_0 R} = \frac{\rho \pi R^2}{2\pi \varepsilon_0 R} = \frac{\rho R}{2 \varepsilon_0}\]Both expressions are equal, which confirms that the electric field is continuous at the boundary \( r = R \).

Step 4: Graph Electric Field as a Function of \( r \)

The electric field inside the cylinder (\( 0 \le r \le R \)) is given by:\[E = \frac{\rho r}{2\varepsilon_0}\]The electric field outside the cylinder (\( r > R \)) is:\[E = \frac{\lambda}{2\pi \varepsilon_0 r} = \frac{\rho R^2}{2 \varepsilon_0 r}\]The graph of \( E \) as a function of \( r \) from \( r=0 \) to \( r=3R \) will start at zero when \( r = 0 \), increase linearly with \( r \) within the cylinder, and then decrease inversely with \( r \) outside the cylinder.

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle used to understand electric fields, particularly around symmetric objects like cylinders. When you place a charge within an enclosed surface, Gauss's Law helps measure the electric flux through that surface. The law states that the total electric flux through a closed surface is equal to the charge enclosed within the surface divided by the permittivity of free space, represented mathematically as:

• \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \)

For a cylinder, this principle allows us to derive expressions for the electric field both inside and outside. By wisely choosing Gaussian surfaces that match the symmetry of the problem, such as a cylindrical shape coaxial with the charge distribution, we simplify the computation of flux and thus find the electric field conveniently.

Charge Density

Charge density, denoted by \( \rho \), describes how charge is distributed over a certain volume. In the context of a charged cylinder, the charge density is uniform, meaning each cubic meter of the cylinder holds the same amount of charge. It plays a critical role when applying Gauss’s Law by specifying the total charge enclosed by a Gaussian surface, calculated as:

• \( Q_{enc} = \rho \times \text{Volume} \)

For cylinders, this volume depends on the radius and length of the Gaussian cylinder used. This total charge is essential for evaluating the electric field strength both inside and outside the cylinder.

Cylindrical Symmetry

Cylindrical symmetry occurs when a problem looks the same from all directions around an axis, much like a round can or a wire. This symmetry simplifies calculations because it means the electric field is the same at any point a fixed distance from the axis. In our scenario, for a uniformly charged solid cylinder, the electric field is calculated using these symmetrical properties. By employing a cylindrical Gaussian surface that matches the symmetry of the cylinder, we can ensure the electric field remains constant over the surface, thereby simplifying the calculations using Gauss's Law.

Electric Field Inside a Cylinder

The electric field inside a uniformly charged cylinder can be derived by considering a Gaussian surface inside the cylinder. The formula derived using Gauss's Law reflects how the electric field strength increases linearly with the distance \( r \) from the axis:

• \( E = \frac{\rho r}{2\varepsilon_0} \)

This linear relationship reflects how the field strengthens as you move away from the center until the surface of the cylinder is reached. This is due to the increasing amount of charge that contributes to the field as the Gaussian cylinder expands within the charged volume.

Electric Field Outside a Cylinder

Outside the charged cylinder, the situation changes. Here, the total charge within the cylinder affects the electric field, and it is useful to express the charge in terms of the charge per unit length \( \lambda \). This relationship emerges from the fact that outside the cylinder, the field is influenced collectively by all the enclosed charge. The electric field decreases with the distance \( r \) from the surface and is given by:

• \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \)

As such, the electric field diminishes in strength as you move farther away, inversely proportional to the distance, highlighting the spreading effect of the field lines emanating from a cylindrical shape. This typically leads to a graph where the field linearly rises from zero and then falls off as you go beyond the boundaries of the charged region.