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# 物理代写|电磁学代写Electromagnetism代考|KYA320 Electrostatic Potential Energy

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## 物理代写|电磁学代写Electromagnetism代考|Electrostatic Potential Energy

Suppose a test charge $q_0$ is placed in the electric field $\mathbf{E}$ of some other charged object. Then, electric force on $q_0$ is $\mathbf{F}=q_0 \mathbf{E}$. If $\mathbf{E}$ is created by a system of charges $q_1, \ldots, q_N$, then from superposition principle
$$\mathbf{E}R=\sum{i=1}^N \mathbf{E}i$$ The electric force acting on $q_0$ is $$\mathbf{F}=\sum{i=1}^N q_0 \mathbf{E}_i=q_0 \mathbf{E}_R$$

Therefore, if the test charge $q_0$ moves in the electric field $\mathbf{E}$, the electrostatic forces (see also Eq. (3.2)) do work on $q_0$.

Suppose that $q_0$ moves in the field $\mathbf{E}$ by some external agent, then the work done by electric field is negative of the work done by external agent. Let $d \mathbf{s}$ be an infinitesimal displacement in the electric field, then work done by the field on test charge $q_0$ is calculated as
$$d W_e=\mathbf{F} \cdot d \mathbf{s}=q_0 \mathbf{E} \cdot d \mathbf{s}=-d U$$
In Eq. (3.3), $-d U$ is the decrease of the potential energy of charge-field system. Therefore,
$$d U=-d W_e=-q_0 \mathbf{E} \cdot d \mathbf{s}$$
That is, the electric field’s work $\mathbf{E}$ decreases electrostatic potential energy of the charge moving in the field.

## 物理代写|电磁学代写Electromagnetism代考|Electric Potential

For a finite displacement of charge from $A$ to some $B$, the change in potential energy of the system charge-field $\Delta U$ is
$$\Delta U=U_B-U_A=\int_A^B d U=-q_0 \int_A^B \mathbf{E} \cdot d \mathbf{s}$$
The integral in Eq. (3.5) is called path integral or line integral, and it is performed along the path that $q_0$ follows as it moves from $A$ to $B$. Since electric force is conservative force, then the integral does not depend on the path taken from $A$ to $B$. The quantity $\frac{U}{q_0}$ is independent of $q_0$, but it depends only on $\mathbf{E}$.
By definition, the ratio $\frac{U}{q_0}$ is called electric potential $\phi$ :
$$\phi=\frac{U}{q_0}$$
Equation (3.6) implies that electric potential, $\phi$, is a scalar quantity.
Potential difference is the difference of electric potential between two points $A$ and $B$ :

\begin{aligned} \Delta \phi &=\phi_B-\phi_A \ &=\frac{\Delta U}{q_0} \ &=-\int_A^B \mathbf{E} \cdot d \mathbf{s} \end{aligned}
Note that potential difference, $\Delta \phi$, is different physical quantity than change in potential energy, $\Delta U$ :
$$\Delta U=q_0 \Delta \phi$$

## 物理代写电磁学代写Electromagnetism代考|Electrostatic Potential Energy

$$\mathbf{E} R=\sum i=1^N \mathbf{E} i$$

$$\mathbf{F}=\sum i=1^N q_0 \mathbf{E}_i=q_0 \mathbf{E}_R$$

$$d W_e=\mathbf{F} \cdot d \mathbf{s}=q_0 \mathbf{E} \cdot d \mathbf{s}=-d U$$

$$d U=-d W_e=-q_0 \mathbf{E} \cdot d \mathbf{s}$$

## 物理代写|电磁学代写Electromagnetism代考|Electric Potential

$$\Delta U=U_B-U_A=\int_A^B d U=-q_0 \int_A^B \mathbf{E} \cdot d \mathbf{s}$$

$$\phi=\frac{U}{q_0}$$

$$\Delta \phi=\phi_B-\phi_A \quad=\frac{\Delta U}{q_0}=-\int_A^B \mathbf{E} \cdot d \mathbf{s}$$

$$\Delta U=q_0 \Delta \phi$$

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