Exercises: Field strength and flux in charge distributions
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…
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.
- a) On which charge do the field lines originate and on which do they terminate?
- 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. a) What happens to the total field at the midpoint of the line segment…
Two equal positive charges are close to each other. Draw their field lines in your notebook.
- 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.
- b) Can two field lines cross at a point? Explain 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.
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…
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.
- a) Reason where the total field points at a point on the axis located outside the segment (analyze which field dominates and its direction).
- 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 dipole is formed by the charges $q_1 = +1\ \text{nC}$ and $q_2 = -1\ \text{nC}$, separated by $20\ \text{cm}$ (take…
A dipole is formed by the charges $q_1 = +1\ \text{nC}$ and $q_2 = -1\ \text{nC}$, separated by $20\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).
- a) Calculate the field (magnitude, direction, and sense) produced by each charge at the midpoint of the line segment joining them.
- b) Calculate the total field at that point.
- c) Compare the result with the field at the midpoint between two equal charges of the same sign and explain the difference.
Two charges $q_1 = +2\ \text{nC}$ and $q_2 = -2\ \text{nC}$ occupy the points $(10, 0)$ and $(-10, 0)$, with coordinates in centimeters (take…
Two charges $q_1 = +2\ \text{nC}$ and $q_2 = -2\ \text{nC}$ occupy the points $(10, 0)$ and $(-10, 0)$, with coordinates in centimeters (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.
- a) Calculate the distance from each charge to point $P$.
- b) Calculate the magnitude of the field produced by each charge at $P$.
- c) Draw the two vectors in your notebook and reason which components cancel each other out and which add up.
- 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 dipole is formed by the charges $q_1 = +2\ \text{nC}$ and $q_2 = -2\ \text{nC}$, separated by $10\ \text{cm}$ (take…
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 center, on the side of the positive charge, is studied.
- a) Determine the distance from the point to each of the charges.
- b) Calculate the field (magnitude and direction) produced by each charge at that point.
- c) Calculate the total field and state its direction.
- 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.
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)). a) Reason,…
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)).
- a) Reason, drawing in your notebook 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. Analyze separately: the axis outside the line segment, the axis between the charges, the perpendicular bisector, and a general point off these lines.
- b) Conclude where the field of a dipole actually vanishes.
- 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…
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.
- a) Reason, by symmetry, that the total field on the perpendicular bisector is perpendicular to the segment joining the charges (which components cancel out?).
- b) Calculate the magnitude of the total field for $y = 0$, for $y = 10\ \text{cm}$ and for $y = 20\ \text{cm}$.
- c) Also calculate the magnitude for $y = 7\ \text{cm}$.
- 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.