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Maxwell’s theory of electromagnetism is a set of equations that describe the behavior of electric and magnetic fields and their interactions with charged particles. The theory provides a unified framework for understanding various phenomena related to electricity and magnetism, such as how charges produce electric fields, how moving charges create magnetic fields, and how changing magnetic fields produce electric fields.
Maxwell’s equations also predict the existence of electromagnetic waves, which are waves of oscillating electric and magnetic fields that can propagate through space at the speed of light. These waves are the basis for many modern technologies, including radio communication, television, and cellular phones.
The importance of Maxwell’s theory of electromagnetism cannot be overstated, as it forms the basis for our understanding of the fundamental workings of many devices and technologies that we rely on in our daily lives.
The line integral of a scalar function $f(x, y, z)$ along a path $C$ is defined as
$$
\int_C f(x, y, z) d s=\lim {\substack{N \rightarrow \infty \ \Delta s_i \rightarrow 0}} \sum{i=1}^N f\left(x_i, y_i, z_i\right) \Delta s_i
$$
where $C$ has been subdivided into $N$ segments, each with a length $\Delta s_i$. To evaluate the line integral, it is convenient to parameterize $C$ in terms of the arc length parameter $s$. With $x=x(s)$, $y=y(s)$ and $z=z(s)$, the above line integral can be rewritten as an ordinary definite integral:
$$
\int_C f(x, y, z) d s=\int_{s_1}^{s_2} f[x(s), y(s), z(s)] d s
$$
As an example, let us consider the following integral in two dimensions:
$$
I=\int_C(x+y) d s
$$
where $C$ is a straight line from the origin to $(1,1)$, as shown in the figure. Let $s$ be the arc length measured from the origin. We then have
$$
\begin{aligned}
& x=s \cos \theta=\frac{s}{\sqrt{2}} \
& y=s \sin \theta=\frac{s}{\sqrt{2}}
\end{aligned}
$$
The endpoint $(1,1)$ corresponds to $s=\sqrt{2}$. Thus, the line integral becomes
$$
I=\int_0^{\sqrt{2}}\left(\frac{s}{\sqrt{2}}+\frac{s}{\sqrt{2}}\right) d s=\sqrt{2} \int_0^{\sqrt{2}} s d s=\left.\sqrt{2} \cdot \frac{s^2}{2}\right|_0 ^{\sqrt{2}}=\sqrt{2}
$$
Use Gauss’s Law to find the direction and magnitude of the electric field in the between the inner and outer cylinders $(a<r<b)$. Express your answer in terms of the total charge $Q$ on the inner cylinder cylinder, the radii $a$ and $b$, the height $l$, and any other constants which you may find necessary.
To use Gauss’s Law, we need to choose a Gaussian surface that encloses the charge distribution. In this case, we can choose a cylindrical Gaussian surface with radius $r$ and height $l$ (same as the height of the cylinders). The axis of the Gaussian surface is aligned with the axis of the cylinders.
By symmetry, the electric field is radial and has the same magnitude at every point on the Gaussian surface. Thus, we can write Gauss’s Law as:
$$\oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$$
where $S$ is the Gaussian surface, $d\mathbf{A}$ is a differential element of area on the surface, $\mathbf{E}$ is the electric field, $Q_{enc}$ is the total charge enclosed by the surface, and $\epsilon_0$ is the permittivity of free space.
The left-hand side of this equation can be simplified to:
$$\oint_S \mathbf{E}\cdot d\mathbf{A} = E(r) \oint_S dA = E(r) \cdot 2\pi rl$$
where $E(r)$ is the magnitude of the electric field at radius $r$, and the integral on the right-hand side is the total area of the cylindrical surface.
The charge enclosed by the Gaussian surface is equal to the charge on the inner cylinder, $Q$, since there is no charge outside the inner cylinder:
$$Q_{enc} = Q$$
Substituting these expressions into Gauss’s Law, we get:
$$E(r) \cdot 2\pi rl = \frac{Q}{\epsilon_0}$$
Solving for the electric field magnitude $E(r)$, we get:
$$E(r) = \frac{Q}{2\pi\epsilon_0 rl}$$
The direction of the electric field is radial, pointing outward from the inner cylinder and inward towards the outer cylinder. Therefore, the electric field points in the $+r$ direction for $a<r<b$, and in the $-r$ direction for $r>b$ and $r<a$.
The voltage difference between the cylinders, $\Delta V$, is defined to be the work done per test charge against the electric field in moving a test charge $q_t$ from the inner cylinder to the outer cylinder
$$
\Delta V \equiv V(a)-V(b)=-\int_b^a \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{s}} \text {. }
$$
Find an expression for the voltage difference between the cylinders in terms of the charge $Q$, the radii $a$ and $b$, the height $l$, and any other constants that you may find necessary.
Let’s assume that the cylinders are coaxial and have a uniform charge distribution with charge density $\rho$.
From Gauss’s Law, the electric field inside a cylinder with uniform charge density $\rho$ is given by:
$E(r)=\frac{\rho}{2 \epsilon_0} r$, for $r \leq a$
where $\epsilon_0$ is the electric constant (permittivity of free space).
Similarly, the electric field outside the cylinder is given by:
$E(r)=\frac{\rho a^2}{2 \epsilon_0 r}$, for $a \leq r \leq b$
Therefore, the voltage difference between the cylinders can be expressed as: \begin{align*} \Delta V &= -\int_b^a \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{s}} \ &= -\int_b^a E(r) dr \ &= -\int_b^a \frac{\rho a^2}{2\epsilon_0 r} dr \ &= \frac{\rho a^2}{2\epsilon_0} \ln{\frac{b}{a}} \end{align*}
where we have used the fact that the integral of $1/r$ is $\ln{r}$.
Now, we can express the charge density $\rho$ in terms of the total charge $Q$ and the volume of the cylinder:
$\rho=\frac{Q}{\pi l\left(b^2-a^2\right)}$
Substituting this expression for $\rho$ in the equation for $\Delta V$, we get: \begin{align*} \Delta V &= \frac{Q}{2\pi\epsilon_0 l} \frac{a^2}{b^2 – a^2} \ln{\frac{b}{a}} \ &= \frac{Q}{2\pi\epsilon_0 l} \ln{\frac{b}{a}} \left(\frac{a}{b+a}\right) \end{align*}
Therefore, the voltage difference between the cylinders is given by $\Delta V = \frac{Q}{2\pi\epsilon_0 l} \ln{\frac{b}{a}} \left(\frac{a}{b+a}\right)$, in terms of the total charge $Q$, the radii $a$ and $b$, the height $l$, and the electric constant $\epsilon_0$.
