Exercises: Field strength and flux in charge distributions

The electric field flux through a flat surface located in a uniform field is defined as $\Phi = E \times S \times \cos \alpha$. Observe the diagram:

1. a) Identify what each symbol in the formula represents, including the angle $\alpha$.
2. b) State the SI unit of flux, constructing it from the units of $E$ and $S$.

A uniform electric field $E = 500\ \text{N/C}$ passes perpendicularly through a flat surface of $0.2\ \text{m}^2$ (the normal to the surface is parallel to the field, $\alpha = 0^\circ$). Calculate the field flux through the surface.

A flat sheet of $0.5\ \text{m}^2$ is inside a uniform electric field $E = 300\ \text{N/C}$, oriented such that the field lines are parallel to the sheet. Observe the diagram:

1. a) What is the angle $\alpha$ between the field and the normal to the sheet?
2. b) Calculate the flux through the sheet and explain, by thinking about the field lines, why that value is obtained.

An electric dipole consists of two equal charges of opposite sign, $+q$ and $-q$, very close to each other; for example, the water molecule behaves like a dipole. Draw the field lines of a dipole in your notebook.

1. a) On which charge do the field lines originate and on which do they terminate?
2. b) Indicate on your drawing a region where the field is strong and another where it is weak, justifying your choice using the density of lines.

You can do it on paper and send a photo with your answer.

Two equal positive charges are close to each other. Draw their field lines in your notebook.

1. a) What happens to the total field at the midpoint of the line segment joining the charges? Relate it to the appearance of your drawing.
2. b) Can two field lines cross at a point? Reason what a crossing would mean for the vector $\vec{E}$ at that point.

You can do it on paper and send a photo with your answer.

A uniform electric field $E = 200\ \text{N/C}$ passes through a flat surface of $0.5\ \text{m}^2$ whose normal forms an angle of $60^\circ$ with the field ($\cos 60^\circ = 0.5$). Calculate the field flux through the surface.

Gauss's law states that the net flux of the electric field through a closed surface depends solely on the net charge enclosed within it. The figure shows an external charge near a closed surface.

1. a) If a closed surface does not enclose any charge, what is the net flux through it?
2. b) Does that mean that the electric field is zero at all points on the surface? Reason your answer by considering a point charge located outside the surface, but very close to it.

Consider a dipole formed by the charges $+q$ and $-q$, and the following points in space: those on the axis joining the charges (outside the line segment) and those on the perpendicular bisector (the line perpendicular to the segment through its midpoint). Draw the field vectors of each charge at those points in your notebook.

1. a) Reason where the total field points at a point on the axis located outside the segment (analyse which field dominates and its sense).
2. b) Reason where the total field points at a point on the perpendicular bisector (hint: look at which components cancel out and which add up).

You can do it on paper and send a photo with your answer.

A cubic box with edge length $a = 10\ \text{cm}$ is immersed in a uniform electric field $E = 3\times10^{3}\ \text{N/C}$, parallel to one of its edges. The box contains no charge. Observe the diagram:

1. a) Calculate the flux through each of the six faces of the cube, stating the sign in each case (convention: flux is positive when lines leave the closed surface and negative when they enter).
2. b) Calculate the net flux through the entire surface of the cube and explain, using Gauss's law, why it has that value.

A dipole is formed by the charges $q_1 = +1\ \text{nC}$ and $q_2 = -1\ \text{nC}$, separated by $20\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Calculate the field (magnitude, direction and sense) produced by each charge at the midpoint of the line segment joining them.
2. b) Calculate the total field at that point.
3. c) Compare the result with the field at the midpoint between two equal charges of the same sign and explain the difference.

A uniform electric field $E = 4\times10^{4}\ \text{N/C}$ passes through a rectangular surface of $30\ \text{cm}$ by $10\ \text{cm}$ whose normal forms an angle of $45^\circ$ with the field ($\cos 45^\circ \approx 0.71$).

1. a) Express the surface area in $\text{m}^2$.
2. b) Calculate the field flux through it.

A point charge $Q = 6\ \text{nC}$ is inside a spherical surface (take $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$).

1. a) Apply Gauss's law, $\Phi = \frac{Q_{int}}{\varepsilon_0}$, and calculate the net flux of the field through the sphere.
2. b) Would the result change if the sphere had twice the radius? Justify your answer.

Two charges $q_1 = +2\ \text{nC}$ and $q_2 = -2\ \text{nC}$ occupy the points $(10, 0)$ and $(-10, 0)$, with coordinates in centimetres (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$; $\cos 45^\circ \approx 0.71$). Point $P(0, 10)$ is studied. The figure shows the geometry of the situation.

1. a) Calculate the distance from each charge to point $P$.
2. b) Calculate the magnitude of the field produced by each charge at $P$.
3. c) Draw the two vectors in your notebook and reason which components cancel each other out and which add up.
4. d) Calculate the magnitude of the total field at $P$ and state its direction and sense.

You can do it on paper and send a photo with your answer.

A point charge $Q = 4\ \text{nC}$ is located outside a closed spherical surface, at a certain distance from it (given: $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$; $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$).

1. a) What is the net flux of the field through the sphere? Apply Gauss's law and justify the result.
2. b) Is the electric field also zero at points on the surface? Reason the difference between net flux and field at each point.
3. c) Calculate the field produced by $Q$ at the point on the sphere closest to it, located at $30\ \text{cm}$ from the charge.

A square frame of $0.1\ \text{m}^2$ can rotate inside a uniform electric field $E = 1000\ \text{N/C}$. The flux is measured for different orientations ($\cos 30^\circ \approx 0.87$; $\cos 60^\circ = 0.5$).

1. a) Calculate the flux when the normal to the frame forms angles of $\alpha = 0^\circ$, $30^\circ$, $60^\circ$ and $90^\circ$ with the field.
2. b) Describe what the graph of flux as a function of $\alpha$ between $0^\circ$ and $180^\circ$ would look like, and reason why flux can turn out to be negative.
3. c) Explain what negative flux means physically, in terms of field lines.

A charge $Q = 3\ \text{nC}$ is enclosed within an irregularly shaped closed surface (for example, the surface of a rucksack), without the charge being at the centre (take $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$).

1. a) Calculate the net field flux through the surface.
2. b) If the charge moves to another position, still inside the surface, does the net flux change? What about the flux through each portion of the surface separately? Reason both answers.

A dipole is formed by the charges $q_1 = +2\ \text{nC}$ and $q_2 = -2\ \text{nC}$, separated by $10\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). A point on the dipole axis located at $50\ \text{cm}$ from its centre, on the side of the positive charge, is studied.

1. a) Determine the distance from the point to each of the charges.
2. b) Calculate the field (magnitude and sense) produced by each charge at that point.
3. c) Calculate the total field and state its sense.
4. d) The two fields almost cancel out, because the distances are similar and the charges are opposite: use this idea to explain why the field of a dipole decreases with distance much faster than that of an isolated charge.

A closed surface encloses the charges $Q_1 = 7\ \text{nC}$ and $Q_2 = -2\ \text{nC}$; a third charge $Q_3 = 8\ \text{nC}$ is located outside the surface (take $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$). Observe the diagram:

1. a) Calculate the net charge enclosed by the surface.
2. b) Calculate the net field flux through it.
3. c) Justify why $Q_3$ does not enter into the calculation, even though it does contribute to the electric field at each point on the surface.
4. d) If $Q_2$ were to become $-7\ \text{nC}$, what would the net flux be? Would that mean that the field is zero on the surface? Reason your answer.

A meter records the electric field flux through a closed surface divided into two regions: through region 1 the flux is $\Phi_1 = -200\ \text{N}\times\text{m}^2/\text{C}$ (the lines enter) and through region 2 it is $\Phi_2 = +350\ \text{N}\times\text{m}^2/\text{C}$ (the lines leave); there are no other regions (take $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$). The figure shows the two regions and the direction of the lines.

1. a) Calculate the net flux through the entire surface.
2. b) Deduce the sign of the net enclosed charge and calculate its value by applying Gauss's law.
3. c) Reason what would happen to the two measured fluxes if the enclosed charge were doubled, with everything else remaining the same.

A hemispherical dome of radius $R = 0.3\ \text{m}$, with its base horizontal, is immersed in a uniform, vertical electric field $E = 400\ \text{N/C}$. Observe the diagram:

1. a) Reason that the flux through the curved surface of the dome is the same as through its horizontal projection (the base circle): the field lines crossing the hemisphere are exactly those crossing its base.
2. b) Calculate this flux.
3. c) If the dome were closed with its flat base, forming a closed surface with no charge inside, what would the net flux be? Explain how the fluxes through the curved part and the base cancel each other out.

Consider a dipole formed by the charges $+q$ and $-q$ (take $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$ for part c)).

1. a) Reason, drawing in your exercise book the field vectors for each charge, whether there is any point in the plane, at a finite distance, where the total field of the dipole is zero. Analyse separately: the axis outside the line segment, the axis between the charges, the perpendicular bisector, and a general point off these lines.
2. b) Conclude where the field of a dipole actually vanishes.
3. c) Calculate the net flux through a closed surface enclosing the whole dipole and relate it to the fact that its field lines originate at $+q$ and terminate at $-q$.

Two identical charges $q = 4\ \text{nC}$ are separated by a distance $d = 20\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). The field at points on the perpendicular bisector of the line segment joining them, at a distance $y$ from the midpoint, is studied. The figure shows the geometry.

1. a) Reason, by symmetry, that the total field on the perpendicular bisector is perpendicular to the segment joining the charges (which components cancel out?).
2. b) Calculate the magnitude of the total field for $y = 0$, for $y = 10\ \text{cm}$ and for $y = 20\ \text{cm}$.
3. c) Also calculate the magnitude for $y = 7\ \text{cm}$.
4. d) The field is zero at $y = 0$ and tends to zero when $y$ is very large, but initially increases with $y$: reason, using the four calculated values, that an intermediate maximum must exist and bound the range of distances in which it lies.

A simulation program draws the electric field line map for a charge distribution: in a region A, $12$ lines cross each $\text{cm}^2$ perpendicular to the field, and in a region B, only $4$ lines. Observe the diagram:

1. a) How are the magnitudes of the field in A and B related? Justify your answer based on the meaning of line density.
2. b) If the field in region B is $200\ \text{N/C}$, calculate the flux through a surface of $0.05\ \text{m}^2$, perpendicular to the field, located in region A.
3. c) The number of lines passing through a surface is proportional to the flux, and lines only originate or terminate where charges are present: use these two ideas to qualitatively explain Gauss's law, that is, why the net flux through a closed surface "counts" the enclosed charge.

You can do it on paper and send a photo with your answer.

A point charge $Q = 4\ \text{nC}$ is at the centre of a spherical surface of radius $R = 0.2\ \text{m}$ (given: $\varepsilon_0 = 8.85\times10^{-12}\ \frac{\text{C}^2}{\text{N}\times\text{m}^2}$; $K = \frac{1}{4 \times \pi \times \varepsilon_0} = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Calculate the net flux through the sphere by applying Gauss's law.
2. b) By symmetry, the field is radial and has the same magnitude at all points on the sphere, so that $\Phi = E \times 4 \times \pi \times R^2$. Rearrange for $E$ and calculate it.
3. c) Calculate the same field using Coulomb's law, $E = K \times \frac{Q}{R^2}$, and verify that it coincides.
4. d) Draw a conclusion about the relationship between Gauss's law and Coulomb's law.