Electric energy and potential

Two point charges $q_1 = 1\,\mu\text{C}$ and $q_2 = 2\,\mu\text{C}$ are separated by $0,3\ \text{m}$ (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Calculate the electric potential energy of the system, $E_p = K \cdot \frac{q_1 \cdot q_2}{r}$.
2. b) Interpret the sign of the result: when separating the charges to infinity, must energy be supplied or is it released? Relate this to the fact that the charges repel each other.

For two protons to fuse inside the Sun, they must overcome their electric repulsion and get extremely close to each other (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$; $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$; $1\ \text{nm} = 10^{-9}\ \text{m}$). The figure shows the potential energy of the pair of protons as a function of their separation.

1. a) Calculate the electric potential energy of two protons separated by a distance $r = 1\ \text{nm}$.
2. b) Express the result in electronvolts.
3. c) Reason why electric repulsion hinders fusion and what role the extremely high temperature in the solar interior plays in overcoming it.

Two charges $q_1 = 2\,\mu\text{C}$ and $q_2 = 3\,\mu\text{C}$, held by insulating supports, go from being separated by $1\ \text{m}$ to being separated by $0,25\ \text{m}$ (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Calculate the initial and final electric potential energy of the system.
2. b) Calculate the work that must be done by an external agent in the process, $W_{ext} = \Delta E_p$.
3. c) Reason the sign of the result: why must work be done to bring together two charges that repel each other?
4. d) What is the work done by the electric field in that same process? Relate it to that of the external agent.

A small sphere in the laboratory has a charge $Q = 3\,\mu\text{C}$ (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Calculate the electric potential at $0,5\ \text{m}$ from the sphere, $V = K \cdot \frac{Q}{r}$.
2. b) State the sign of the potential and reason where, by convention, $V = 0$ is taken.

Electric potential is a scalar quantity; electric field, a vector quantity.

1. a) Explain the practical difference this makes when adding up the contributions from several charges at a point.
2. b) Can the potential at a point be negative? Give an example.
3. c) What sign does the potential created by a positive charge have?

Two charges $q_1 = 4\ \text{nC}$ and $q_2 = -2\ \text{nC}$ are at $0,3\ \text{m}$ and $0,5\ \text{m}$, respectively, from a point P (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Calculate the potential created by each charge at P.
2. b) Calculate the total potential at P.
3. c) Explain why here the contributions are added directly as numbers, whereas for the electric field vectors must be added.

A charge $q = 5\,\mu\text{C}$ moves from a point A, with potential $V_A = 200\ \text{V}$, to a point B, with $V_B = 80\ \text{V}$.

1. a) Calculate the work done by the electric field, $W_{A \to B} = q \cdot (V_A - V_B)$.
2. b) Reason, from the sign of the result, whether the field spontaneously "pushes" the charge in that direction of displacement.

A charge $q = -3\,\mu\text{C}$ moves from a point A ($V_A = 500\ \text{V}$) to a point B ($V_B = 100\ \text{V}$).

1. a) Calculate the work done by the field, $W_{A \to B} = q \cdot (V_A - V_B)$.
2. b) Interpret the negative sign of the result: is it a spontaneous movement or must it be carried out by an external agent?
3. c) Reason where this negative charge would move spontaneously: towards higher potentials or towards lower potentials?

In a region, the equipotential surfaces are parallel planes. Two of them, separated by $2\ \text{cm}$, have potentials $V_1 = 300\ \text{V}$ and $V_2 = 100\ \text{V}$. Observe the diagram:

1. a) Calculate the electric field between the two planes, assuming it to be uniform, $E = \frac{\Delta V}{d}$, and state its direction and sense (the field points towards decreasing potentials).
2. b) Calculate the work done by the field when moving a charge $q = 2\ \text{nC}$ from the plane of $300\ \text{V}$ to that of $100\ \text{V}$.
3. c) A student moves the charge first "sideways", without leaving the plane of $300\ \text{V}$, and only then carries it to the plane of $100\ \text{V}$ along a broken path. Reason, using the fact that the electric field is conservative (work depends only on the initial and final points), that the total work is the same as in part b), and calculate the work for the "sideways" section.

In particle physics, the electronvolt is widely used: $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$ (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$).

1. a) Express an energy of $3\ \text{eV}$ in joules.
2. b) Justify the equivalence above by calculating, using $W = q \cdot \Delta V$, the energy gained by an electron when accelerated through a potential difference of $1\ \text{V}$.

In an electron microscope, an electron starts from rest and is accelerated through a potential difference $\Delta V = 2000\ \text{V}$ (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; electron mass $m_e = 9,11\cdot10^{-31}\ \text{kg}$). Observe the diagram:

1. a) Calculate the kinetic energy gained by the electron, in joules and in electronvolts.
2. b) Apply conservation of energy, $\frac{1}{2} \cdot m_e \cdot v^2 = e \cdot \Delta V$, and calculate its final speed.

In a hospital X-ray tube, electrons are accelerated towards the anode with a potential difference $\Delta V = 40\ \text{kV}$ (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; electron mass $m_e = 9,11\cdot10^{-31}\ \text{kg}$; speed of light $c = 3\cdot10^{8}\ \text{m/s}$). The figure shows the diagram of the tube.

1. a) Express the potential difference in volts and calculate the kinetic energy of each electron on reaching the anode, in electronvolts and in joules.
2. b) Calculate the final speed of the electron using the classical expression $\frac{1}{2} \cdot m_e \cdot v^2 = e \cdot \Delta V$.
3. c) Compare the result with the speed of light and comment on why, at these energies, the classical approximation starts to become questionable.

Two identical charges $q = 5\ \text{nC}$ occupy the points $(0, 0)$ and $(0, 40)$, with coordinates in centimetres (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Reason, drawing the vectors in your exercise book, that the total electric field at the midpoint $P(0, 20)$ is zero.
2. b) Calculate the electric potential at that same point P.
3. c) Draw a conclusion: is it true that where the field is zero, the potential is also zero?

Your classmate claims: "if at a point the electric field is zero, the potential is also zero; and if the potential is zero, the field is also zero" (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

1. a) Refute the first part with a numerical example: find the point on the line segment between the charges $q_1 = 9\ \text{nC}$ and $q_2 = 4\ \text{nC}$, separated by $1\ \text{m}$, where the field is zero, and calculate the potential at that point.
2. b) Refute the second part with another numerical example: calculate the potential and the field at the midpoint between the charges $q_1 = +5\ \text{nC}$ and $q_2 = -5\ \text{nC}$, separated by $40\ \text{cm}$.
3. c) Explain, from the definitions of field (vector) and potential (scalar), why both statements are false in general.

During a storm, the base of a cloud can be at a potential of about $10^{8}\ \text{V}$ relative to the ground, and a lightning bolt can transfer a charge of $20\ \text{C}$ (given: $1\ \text{kW}\cdot\text{h} = 3,6\cdot10^{6}\ \text{J}$).

1. a) Calculate the energy released in the lightning bolt, $E = Q \cdot \Delta V$.
2. b) Express it in kilowatt-hours.
3. c) The monthly electricity consumption of a household is about $200\ \text{kW}\cdot\text{h}$: compare both values and reason why, despite this huge amount of energy, it is so difficult to harness lightning energy (think about the instantaneous power and the unpredictability of the discharges).