### 物理代写|电动力学代写electromagnetism代考|ELEC3104

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• Foundations of Data Science 数据科学基础

## 物理代写|电动力学代写electromagnetism代考|ELECTRIC POTENTIAL

We often come across the term potential when applied to the potential energy of a body or the potential difference between two points in a circuit. In the former case, the potential energy of a body is related to its height above a certain reference level. Thus, a body gains potential energy when we raise it to a higher level. This gain in energy is equal to the work done against an attractive force, gravity in this example. Figure 2.6a shows this situation.

As Figure 2.6a shows, the body is placed in an attractive, gravitational force field. So, if we raise the body through a certain distance, we have to do work against the gravitational field. The difference in potential energy between positions 1 and 2 is equal to the work done in moving the body from 1 to 2 , a distance of $l$ metres. This work done is given by
$$F \times l=m \times 9.81 \times l$$

where $m$ is the mass of the body $(\mathrm{kg})$ and $9.81$ is the acceleration due to gravity $\left(\mathrm{m} \mathrm{s}^{-2}\right.$ ). (Although the effects of gravity vary according to the inverse square law, the difference in gravitational force between positions 1 and 2 is small. This is because the Earth is so large. Thus, we can take the gravitational field to be linear in form, and so this equation holds true.)

In an electrostatic field, we have an electrostatic force field instead of a gravitational force field. However, the idea of potential energy is the same. Let us consider the situation in Figure 2.6b. We have a positive test charge of $1 \mathrm{C}$ at a distance $\mathrm{d}_1$ from the fixed negative charge, $-q_1$. This test charge will experience an attractive force whose magnitude we can find from Coulomb’s law. Now, if we move the test charge from position 1 to position 2, we have to do work against the field. If the distance between positions 1 and 2 is reasonably large, the strength of the force field decreases as we move away from the fixed charge. Thus, we say that we have a non-linear field.
As the field decreases when we move away from the fixed charge, let us move the test charge a very small distance, $\mathrm{d} r$. The electric field strength will hardly alter as we move along this small distance. So, the work done against the field in moving the test charge a small distance $\mathrm{d} r$ will be given by
\begin{aligned} \text { work done } &=\text { force } \times \text { distance } \ &=-F \times \mathrm{d} r \ &=-1 \times E \times \mathrm{d} r \end{aligned}

## 物理代写|电动力学代写electromagnetism代考|EQUIPOTENTIAL LINES

Let us consider the three paths $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ shown in Figure $2.7 \mathrm{a}$. All of these paths link points 1 and 2, but only path A does so directly. Now, let us take the circular lines in Figure 2.7a as the contours on a hill. In moving from position 1 to position 2 by way of path $\mathrm{A}$, we clearly do work against gravity. The work done is equal to the gain in potential energy which, in turn, is equal to the gravitational force times the change in vertical height. (This is shown in Figure 2.7b.)

Now let us take path B. We initially walk left from position 1, around the contour line, to a point directly below position 2. As we have moved around a contour line, we have not gained any height, and so the potential energy remains the same, i.e., we have not done any work against gravity. We now have to walk uphill to position 2 . In doing so we do work against gravity equal to the gain in potential energy. This gain in potential energy is clearly the same as with path $\mathrm{A}$. (Although we have to do more physical work in travelling along path $\mathrm{B}$, the change in potential energy is the same.) If we use path $\mathrm{C}$, the same argument holds true. So, we can say that the work done against gravity is independent of the path we take.

Let us now turn our attention to the electrostatic field in Figure 2.8. As with the contour map, we have three different paths. As we have just seen, we do no work against the field when we move in a circular direction. We only do work when we move in a radial direction. Thus, the potential difference between points 1 and 2 is independent of the exact path we take. This implies that we do no work against the field when we move around the plot in a circular direction. Thus, the circular ‘contours’ in Figure $2.8$ are lines of equal potential or equipotential lines.

We should be careful when using the term equipotential lines. This is because we are considering a point charge, and so the equipotential surfaces are actually spheres with the charge at their centre. As we are not yet able to draw in a three-dimensional holographic world, we have to make do with two-dimensional diagrams drawn on pieces of paper!

## 物理代写|电动力学代写electromagnetism代考|ELECTRIC POTENTIAL

$$F \times l=m \times 9.81 \times l$$

$$\text { work done }=\text { force } \times \text { distance } \quad=-F \times \mathrm{d} r=-1 \times E \times \mathrm{d} r$$

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## MATLAB代写

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