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+# 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 |
+\ No newline at end of file