# Test 4 – Answers

We assume electrostatic fields in a vacuum and use:

$$
k=\frac{1}{4\pi\varepsilon_0}.
$$

Here, $U$ denotes electric potential energy, $V$ denotes electric potential, and $q_0$ is the charge being moved. 4Te_pw14 (1)(1)

## 1. Work done by the electric force from $a$ to $b$

The work is found by integrating the force along the path of the charge:

$$
W_{ab}=\int_a^b\vec F_e\cdot d\vec r.
$$

Since $\vec F_e=q_0\vec E$, we obtain:

$$
\boxed{W_{ab}=q_0\int_a^b\vec E\cdot d\vec r}.
$$

The electrostatic force is conservative. The work therefore equals the decrease in potential energy:

$$
\boxed{W_{ab}=U_a-U_b=q_0(V_a-V_b)}.
$$

Positive work means that the potential energy decreases.

## 2. Condition for conservation of energy

The total energy of an **isolated system** is constant:

$$
\boxed{\Delta E_{\mathrm{total}}=0}.
$$

If only the electric force does work, the sum of kinetic and electric potential energy is conserved:

$$
\boxed{K_a+U_a=K_b+U_b}.
$$

For a particle of mass $m$:

$$
\frac12mv_a^2+U_a=\frac12mv_b^2+U_b.
$$

This follows from the work–energy theorem:

$$
\Delta K=W_{ab}=-\Delta U.
$$

## 3. Motion in the field of a point charge

### 3a. Derivation of the work

A fixed point charge $q$ creates the field:

$$
\vec E(r)=k\frac{q}{r^2}\hat r.
$$

The force on $q_0$ is:

$$
\vec F_e=k\frac{qq_0}{r^2}\hat r.
$$

Let $r_a$ and $r_b$ be the distances from $q$ to $a$ and $b$, respectively. Since $\hat r\cdot d\vec r=dr$, we obtain:

$$
W_{ab}
=kqq_0\int_{r_a}^{r_b}\frac{1}{r^2}\,dr.
$$

Thus:

$$
W_{ab}
=kqq_0\left[-\frac1r\right]_{r_a}^{r_b}.
$$

Therefore:

$$
\boxed{W_{ab}=kqq_0\left(\frac1{r_a}-\frac1{r_b}\right)}.
$$

The result depends only on the initial and final distances, not on the path taken.

### 3b. Potential energy with a reference at infinity

We choose:

$$
U(\infty)=0.
$$

As $r_b\to\infty$, the work becomes:

$$
W_{a\infty}=\frac{kqq_0}{r_a}.
$$

Since $W_{a\infty}=U_a-U_\infty$, we obtain:

$$
\boxed{U_a=\frac{kqq_0}{r_a}},
\qquad
\boxed{U_b=U_\infty=0}.
$$

The general expression for the potential energy is therefore:

$$
\boxed{U(r)=\frac{kqq_0}{r}}.
$$

Charges with the same sign give positive potential energy; charges with opposite signs give negative potential energy.

## 4. Assembling point charges

Let $r_{ij}$ be the distance between charges $q_i$ and $q_j$. The zero of potential energy is chosen to be when all charges are infinitely far apart.

### 4a. Total potential energy of $q_1,q_2,q_3$

The charges can be assembled one at a time:

- $q_1$ requires no work, since no other charges are present yet.
- $q_2$ adds the energy $kq_1q_2/r_{12}$.
- $q_3$ adds the energy $kq_1q_3/r_{13}+kq_2q_3/r_{23}$.

The total potential energy is:

$$
\boxed{
U_{123}
=k\left(
\frac{q_1q_2}{r_{12}}
+\frac{q_1q_3}{r_{13}}
+\frac{q_2q_3}{r_{23}}
\right)
}.
$$

Each pair of charges is counted once.

### 4b. Additional potential energy when adding $q_0$

With the three existing charges held in place, adding $q_0$ gives:

$$
\boxed{
\Delta U
=k\left(
\frac{q_0q_1}{r_{01}}
+\frac{q_0q_2}{r_{02}}
+\frac{q_0q_3}{r_{03}}
\right)
}.
$$

This can also be written as:

$$
\boxed{\Delta U=q_0V_{123}},
$$

where $V_{123}$ is the potential due to $q_1,q_2,q_3$ at the position where $q_0$ is placed.

## 5. Motion in a uniform electric field

### 5a. Derivation of the work

In a uniform field, $\vec E$ is constant:

$$
W_{ab}=q_0\int_a^b\vec E\cdot d\vec r
=q_0\vec E\cdot(\vec r_b-\vec r_a).
$$

Thus:

$$
\boxed{W_{ab}=q_0\vec E\cdot(\vec r_b-\vec r_a)}.
$$

If the displacement has magnitude $s$ and the angle between the displacement and the field is $\alpha$:

$$
\boxed{W_{ab}=q_0Es\cos\alpha}.
$$

Choosing the $x$-axis along the direction of the field gives:

$$
\boxed{W_{ab}=q_0E(x_b-x_a)}.
$$

### 5b. Potential energy at $a$ and $b$

Since $W_{ab}=U_a-U_b$:

$$
U_b-U_a=-q_0E(x_b-x_a).
$$

A potential energy that satisfies this is:

$$
U(x)=-q_0Ex+C.
$$

Therefore:

$$
\boxed{U_a=-q_0Ex_a+C},
\qquad
\boxed{U_b=-q_0Ex_b+C}.
$$

The constant $C$ is determined by the chosen zero of potential energy. Choosing $U_a=0$ gives:

$$
\boxed{U_a=0},
\qquad
\boxed{U_b=-q_0E(x_b-x_a)}.
$$

For a positive charge, the potential energy decreases when it moves in the direction of the field.

## 6. Electric dipole in a uniform electric field

A dipole consists of charges $+q$ and $-q$, separated by the vector $\vec d$, which points from the negative charge to the positive charge.

The dipole moment is defined as:

$$
\boxed{\vec p_e=q\vec d}.
$$

The angle between $\vec p_e$ and $\vec E$ is denoted by $\theta$.

### 6a. Derivation of the torque

The forces on the charges are:

$$
\vec F_+=q\vec E,
\qquad
\vec F_-=-q\vec E.
$$

The net force is zero, but the forces form a couple.

Taking the midpoint of the dipole as the origin, the position vectors are:

$$
\vec r_+=\frac{\vec d}{2},
\qquad
\vec r_-=-\frac{\vec d}{2}.
$$

The total torque is:

$$
\begin{aligned}
\vec\tau
&=\vec r_+\times\vec F_+
+\vec r_-\times\vec F_-\\
&=\frac{\vec d}{2}\times q\vec E
+\left(-\frac{\vec d}{2}\right)\times(-q\vec E)\\
&=q\vec d\times\vec E.
\end{aligned}
$$

Thus:

$$
\boxed{\vec\tau=\vec p_e\times\vec E}.
$$

The magnitude of the cross product gives:

$$
\boxed{\tau=p_eE\sin\theta}.
$$

The torque tends to align the dipole moment with the field.

### 6b. Work during rotation from $\theta_a$ to $\theta_b$

The torque acts in the direction of decreasing $\theta$. With $\tau=p_eE\sin\theta$ as its magnitude:

$$
dW=-\tau\,d\theta.
$$

The work is therefore:

$$
W_{ab}=-p_eE\int_{\theta_a}^{\theta_b}\sin\theta\,d\theta.
$$

Integration gives:

$$
\boxed{
W_{ab}=p_eE(\cos\theta_b-\cos\theta_a)
}.
$$

Since $W_{ab}=U_a-U_b$, the potential energy can be written as:

$$
U(\theta)=-p_eE\cos\theta+C.
$$

Therefore:

$$
\boxed{U_a=-p_eE\cos\theta_a+C},
$$

$$
\boxed{U_b=-p_eE\cos\theta_b+C}.
$$

### 6c. Reference at $90^\circ$ and the general expression

The zero of potential energy can be chosen freely. We choose:

$$
\boxed{U_a=U(90^\circ)=0}.
$$

At $90^\circ$, the two charges are at the same potential in the uniform field, so their contributions to the total potential energy cancel each other.

Since $\cos90^\circ=0$, this choice of reference gives $C=0$. Therefore:

$$
\boxed{U(\theta)=-p_eE\cos\theta}.
$$

Using the dot product, we obtain:

$$
\boxed{U=-\vec p_e\cdot\vec E}.
$$

| Angle | Orientation | Potential energy | Equilibrium |
|---|---|---|---|
| $0^\circ$ | Dipole moment aligned with the field | $-p_eE$, minimum | Stable |
| $90^\circ$ | Dipole moment perpendicular to the field | $0$ | No equilibrium |
| $180^\circ$ | Dipole moment opposite to the field | $+p_eE$, maximum | Unstable |