# 电、磁和波代写|Electricity, Magnetism and Waves PX1221 Cardiff University Assignment

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## Instructions:

Electricity, magnetism, and waves are three interconnected branches of physics that deal with the behavior and properties of electromagnetic fields and phenomena.

Electricity refers to the flow of electric charge, usually carried by electrons in a conductor, and is characterized by voltage, current, and resistance. It is essential to many aspects of modern life, from powering our homes and electronic devices to driving industry and transportation.

Magnetism refers to the behavior of magnets and magnetic fields, which are produced by moving electric charges or certain materials called ferromagnetic materials. Magnetic fields are important in many applications, including electric motors, generators, and medical imaging.

Waves refer to the propagation of energy through space, which can be in the form of electromagnetic waves or mechanical waves like sound waves. Electromagnetic waves, such as radio waves, microwaves, light waves, and X-rays, are characterized by their wavelength, frequency, and amplitude and play a vital role in communication, broadcasting, and imaging.

The study of electricity, magnetism, and waves is fundamental to understanding many aspects of the physical world, from the behavior of atoms and molecules to the functioning of modern technology.

(a) Find the total charge $Q$ on the rectangular surface of length $a$ ( $x$ direction from $x=0$ to $x=a$ ) and width $b(y$ direction from $y=0$ to $y=b)$, if the charge density is $\sigma(x, y)=k x y$, where $k$ is a constant.

To find the total charge $Q$ on the rectangular surface, we need to integrate the charge density $\sigma(x, y)$ over the surface area:

$$Q = \int_{0}^{b} \int_{0}^{a} \sigma(x, y) dx dy$$

Substituting $\sigma(x,y)=kxy$ into the above expression and performing the integrals yields:

\begin{align*} Q &= \int_{0}^{b} \int_{0}^{a} k x y dx dy \ &= k \int_{0}^{b} \left[\frac{x^2}{2} y \right]{x=0}^{x=a} dy \ &= k \int{0}^{b} \frac{a^2}{2} y dy \ &= \frac{k a^2}{2} \left[\frac{y^2}{2} \right]_{y=0}^{y=b} \ &= \frac{k a^2 b^2}{4} \end{align*}

Therefore, the total charge on the rectangular surface is $\boxed{\frac{k a^2 b^2}{4}}$.

(b) Find the total charge on a circular plate of radius $R$ if the charge distribution is $\sigma(r, \theta)=k r(1-\sin \theta)$.

To find the total charge on the circular plate of radius $R$ with charge distribution $\sigma(r,\theta) = kr(1-\sin\theta)$, we need to integrate the charge density over the entire area of the plate.

The charge density $\sigma$ is a function of $r$ and $\theta$, where $r$ is the radial distance from the center of the plate and $\theta$ is the angle measured from some reference direction. Since the plate is circular, we can integrate over $r$ and $\theta$ as follows:

\begin{align*} Q &= \int_{0}^{2\pi}\int_{0}^{R} \sigma(r,\theta)r,dr,d\theta \ &= \int_{0}^{2\pi}\int_{0}^{R} k r^2 (1-\sin\theta),dr,d\theta\ &= k \int_{0}^{2\pi}\left[\frac{r^3}{3} – r^2\cos\theta\right]{r=0}^{r=R},d\theta\ &= k \int{0}^{2\pi}\left(\frac{R^3}{3} – R^2\cos\theta\right),d\theta\ &= k\left[\frac{R^3}{3}\theta – R^2\sin\theta\right]_{\theta=0}^{\theta=2\pi}\ &= k\left(\frac{2}{3}\pi R^3 – 0\right)\ &= \frac{2}{3}\pi k R^3 \end{align*}

Therefore, the total charge on the circular plate is $\frac{2}{3}\pi k R^3$.

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 apply Gauss’s Law, we need to choose a closed surface that encloses the region of interest, which in this case is the region between the inner and outer cylinders. A convenient choice is a cylindrical Gaussian surface of radius $r$ and height $l$, centered on the axis of the cylinders. The electric field will be perpendicular to the surface at every point, so the flux of the electric field through the surface is simply the product of the magnitude of the field and the area of the curved surface:

$$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = E(r) 2\pi rl$$

where $E(r)$ is the magnitude of the electric field at radius $r$ and $d\vec{A}$ is a differential area element of the surface.

By Gauss’s Law, the flux through the surface is equal to the enclosed charge divided by the permittivity of free space:

$$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$$

where $Q_{enc}$ is the charge enclosed by the Gaussian surface. Since the inner cylinder is the only source of charge inside the surface, the enclosed charge is simply $Q_{enc}=Q$.

Setting these two equations equal to each other, we have

$$E(r) 2\pi rl = \frac{Q}{\epsilon_0}$$

Solving for $E(r)$, we obtain:

$$E(r) = \frac{Q}{2\pi\epsilon_0 rl}$$

The direction of the electric field is radial, i.e., pointing outward from the axis of the cylinders if $Q$ is positive and inward if $Q$ is negative.

Therefore, the magnitude of the electric field between the cylinders is given by:

$$\boxed{E(r) = \frac{Q}{2\pi\epsilon_0 rl}}$$