Question

A hollow, thin-walled insulating cylinder of radius $R$ and length $L$ (like the cardboard tube in a roll of toilet paper) has charge $Q$ uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if $L\ll R$, the result of part (a) reduces to the potential on the axis of a ring of charge of radius $R$. (See Example 23.11 in Section 23.3.) (c) Use the result of part (a) to find the electric field at all points along the axis of the tube.

Short answer

Potential on axis: $V=\frac{kQ}{L}\ln \left(\frac{z+\sqrt{R^2+z^2}+L/2}{z+\sqrt{R^2+z^2}-L/2}\right)$; if $L\ll R$, it simplifies to a ring's potential. Electric field: $E=-\frac{dV}{dz}$.

Step-by-step solution

Identify the setup of the problem

We have a hollow, thin-walled insulating cylinder with charge $Q$ uniformly distributed over its surface. The cylinder has radius $R$ and length $L$. We need to find the electric potential along the axis of the tube, assuming potential is zero at infinity.

Derive the expression for the potential along the axis (Part a)

Considering a small charge element $dq$ on the surface of the cylinder, at a distance $z$ from the point along the axis where the potential is being calculated. The linear charge density $\lambda =\frac{Q}{L}$. For a point at distance $z$ along the axis, the potential $dV$ due to $dq$ is $dV=\frac{k\cdot dq}{\sqrt{R^2+z^2}}$. Integrate over the length of the cylinder (from $-\frac{L}{2}$ to $\frac{L}{2}$): $$V={\int }_{-\frac{L}{2}}^{\frac{L}{2}}\frac{k\lambda Rd\theta }{\sqrt{R^2+z^2}}=\frac{kQ}{L}{\int }_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{\sqrt{R^2+z^2}}dz.$$ Solving this integral, we find: $$V=\frac{kQ}{L}\ln \left(\frac{z+\sqrt{R^2+z^2}+\frac{L}{2}}{z+\sqrt{R^2+z^2}-\frac{L}{2}}\right).$$

Simplify for $L\ll R$ and compare with a ring of charge (Part b)

If $L\ll R$, the cylinder approximates a ring of charge, and the length $L$ becomes negligible compared to $R$. The expression simplifies to the potential on axis of a ring of charge:$$V=\frac{kQ}{\sqrt{R^2+z^2}}.$$ This matches the expected formula for a ring of charge.

Derive the electric field along the axis (Part c)

The electric field is the negative gradient of potential, so we take the derivative of the potential $V$ with respect to $z$. Using the earlier derived potential: $$E=-\frac{dV}{dz}=-\frac{kQ}{L}(\frac{1}{\sqrt{R^2+(z+\frac{L}{2}{)}^2}}-\frac{1}{\sqrt{R^2+(z-\frac{L}{2}{)}^2}}).$$ This gives the electric field along the axis of the cylinder.

Key concepts

Electric Field

The electric field is a fundamental concept in physics, representing the force that a charged object exerts on any other charge in its vicinity. Think of it like an invisible web of influence surrounding a charge.
If you place another charge within this web, that second charge will experience a force without actually touching the first charge. This field is represented by vectors that illustrate the direction and magnitude of the force felt by a positive test charge.

Mathematically, the electric field $E$ can be derived from the electric potential $V$. It is related to the potential by the gradient where $$E=-\frac{dV}{dz}\text{ (for one-dimensional motion)}.$$ It signifies that the electric field direction is steepest in the direction where potential decreases fastest. When analyzing fields, especially for geometrically symmetric objects like cylinders, the field typically points perpendicular to surfaces and is strongest where lines are closest together. This way, calculating and understanding the potential helps find the electric field in complex arrangements, like those of cylinders of charge.

Cylinder of Charge

A cylinder of charge is a common model in electrostatics. Imagine a long hollow tube that has charge spread evenly over its surface. This setup helps when studying fields and potentials due to symmetrical charge distributions.
With a cylinder, variables like radius $R$, length $L$, and total charge $Q$ matter tremendously. These parameters define how charges are spread and influence points within or outside the cylinder.

Working with cylinders simplifies complex geometrical models. By breaking the problem into smaller components, like infinitesimally small charge elements, students can use integration to determine forces, fields, and potentials under various conditions.

Uniform Charge Distribution

Uniform charge distribution means that each piece of the charged object carries the same amount of charge per unit area or length. It's like spreading peanut butter smoothly over a slice of bread - every part of the bread gets an equal amount of peanut butter. In our hollow cylinder, this uniform spread means that each small segment of the surface carries $\lambda =\frac{Q}{L}$, where $\lambda$ is the linear charge density.

This uniformity ensures that calculations for potential and electric fields can take advantage of symmetry. This leads to simpler integrals and more straightforward math. Without it, finding fields or potentials would be more complex and less likely to have elegant solutions. Hence, understanding uniform distribution is key to efficiently solving electrostatic problems.

Integral Calculus

Integral calculus is crucial when dealing with continuous charge distributions, like our thin-walled cylinder. It's a part of calculus focused on accumulation, telling us the total effect of adding up many small pieces, which is essential for electromagnetism.
By integrating over a charge distribution, one can find the total potential or field at a point. For a cylinder of charge, this involves integration over each element of the cylinder, considering how each affects a point on its axis.

The basic principle: break down the charge into infinitesimal parts $dq$, compute their individual potentials, and then sum these. For instance, the integration for potential involves the equation:
$$V=\int \frac{k\cdot dq}{\sqrt{R^2+z^2}}$$ which tells us how the entire charged cylinder influences a point along its axis.

Ring of Charge Approximation

When a cylinder's length $L$ is much smaller than its radius $R$ ($L\ll R$,), it can be approximated as a ring of charge. This is because the length becomes negligibly small compared to how far the charge reaches around the cylindrical curve.
Approximating the cylinder as a ring simplifies calculations since evaluating the potential or field from a ring is significantly easier due to its inherent symmetry.

• The potential for a ring is derived more simply as:
  $$V=\frac{kQ}{\sqrt{R^2+z^2}}$$ which illustrates how potent the symmetry of a ring can be in simplifying electrostatic calculations.

This approach is a powerful tool to make complex problems more tractable by trading off some precision for conceptual simplicity, making it a cornerstone in physics problem-solving.