Exercises: Vector treatment: free charges in fields

A test charge $q = 2\,\mu\text{C}$, placed at a point in a laboratory apparatus, experiences an electric force of $0.06\ \text{N}$. The situation is shown in the figure.

1. a) Calculate the electric field strength at that point, $E = \frac{F}{q}$, and state its SI unit.
2. b) Since the charge is positive, how are the direction and sense of $\vec{E}$ related to those of the force $\vec{F}$?

To measure the electric field at a point, a test charge is used.

1. a) Explain what a test charge is and why it should have a small value, so that it does not disturb the charges creating the field.
2. b) Write down the expression for the field created by a point charge $Q$ at a distance $r$ and reason, from it, why the field does not depend on the test charge used to measure it.

A small sphere charged with $Q = 5\,\mu\text{C}$ creates an electric field around it (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). Look at the diagram:

1. a) Calculate the magnitude of the field at $0.3\ \text{m}$ from the sphere.
2. b) State the direction and sense of the field at that point, reasoning from the sign of the charge.

Between the plates of a capacitor there is a uniform electric field $\vec{E} = 3\cdot10^{4}\,\vec{j}\ \text{N/C}$. A particle with charge $q = -4\,\mu\text{C}$ is placed inside it, as shown in the figure.

1. a) Calculate the electric force vector acting on the particle, $\vec{F} = q \cdot \vec{E}$, expressing it using the corresponding unit vector.
2. b) Explain why, since the charge is negative, the force is in the opposite direction to the field.

A space probe accumulates a net charge of $2\ \text{nC}$ due to bombardment by solar particles (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). Assume the probe is approximately point-like.

1. a) Calculate the magnitude of the electric field it creates at $10\ \text{cm}$ from its surface (express the distance in meters).
2. b) Also calculate the field at $1\ \text{m}$ and check by what factor it has decreased when multiplying the distance by ten.

Draw in your notebook the electric field lines produced by a positive point charge and, separately, those of a negative point charge.

1. a) What does the direction of the arrows on the lines indicate? Relate it to the force that a positive test charge would experience.
2. b) Where is the field most intense, according to your drawing? Relate it to the expression $E = K \times \frac{Q}{r^2}$.

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

The field produced by a point charge is $E$ at a distance $r$.

1. a) Using the expression $E = K \times \frac{Q}{r^2}$, reason out what value the field will have at a distance $2 \times r$ and at a distance $\frac{r}{3}$.
2. b) If the field at $0.5\ \text{m}$ from a certain charge is $7200\ \text{N/C}$, calculate its value at $1\ \text{m}$ without using the constant $K$.

A sensor measures an electric field of $4500\ \text{N/C}$ at $2\ \text{m}$ from the center of a charged metal sphere in the lab; the field points toward the sphere (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). The situation is represented in the figure.

1. a) Rearrange the expression $E = K \times \frac{Q}{r^2}$ to solve for charge and calculate its absolute value.
2. b) From the direction of the measured field, reason what the sign of the charge on the sphere is.

In a cathode ray tube, an electron passes through a region with a uniform electric field $E = 1000\ \text{N/C}$ (given: elementary charge $e = 1.6\times10^{-19}\ \text{C}$; mass of the electron $m_e = 9.11\times10^{-31}\ \text{kg}$).

1. a) Calculate the electric force on the electron.
2. b) Calculate its acceleration by applying Newton's second law, $a = \frac{F}{m_e}$.

A point charge $Q = 4\ \text{nC}$ is at the origin of coordinates (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). Observe the diagram:

1. a) Calculate the magnitude of the electric field at the point $P(3, 4)$, with the coordinates in meters.
2. b) The radial unit vector at that point is $\vec{u}_r = \frac{3}{5}\,\vec{i} + \frac{4}{5}\,\vec{j}$. Write the vector $\vec{E}$ in terms of $\vec{i}$ and $\vec{j}$.

Two point charges $q_1 = 3\,\mu\text{C}$ and $q_2 = -2\,\mu\text{C}$ lie on the $x$-axis, at $x_1 = 0$ and $x_2 = 1\ \text{m}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). The figure shows the arrangement of the charges.

1. a) Calculate the field (magnitude and direction) produced by each charge at the point $x = 3\ \text{m}$.
2. b) Apply the superposition principle and calculate the total field at that point, stating its direction and sense.

Two identical charges $q = 5\ \text{nC}$ are separated by $40\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) By drawing the vectors in your notebook, reason what the value of the total field is at the midpoint of the line segment joining the charges.
2. b) Calculate the total field (magnitude, direction, and sense) at a point on the segment located at $10\ \text{cm}$ from one of the charges.

A charge $Q = 8\ \text{nC}$ is fixed to an insulating stand in the lab (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Calculate the electric field (magnitude and direction) at $25\ \text{cm}$ from the charge.
2. b) Calculate the force (magnitude and direction) that a charge $q_0 = 3\,\mu\text{C}$ placed at that point would experience, using $\vec{F} = q_0 \cdot \vec{E}$.

The manual of a medical diagnostic instrument requires that the electric field at its location must not exceed $500\ \text{N/C}$. Equipment that accumulates a point charge of $2\ \text{nC}$ operates in the same room (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Rearrange the expression $E = K \times \frac{Q}{r^2}$ to solve for distance, and calculate the minimum distance from the equipment at which the instrument must be installed.
2. b) If it is ultimately installed at twice that distance, by what factor is the field reduced? Explain your reasoning.

A proton is between the plates of a capacitor, where the electric field is uniform and equal to $E = 2\times10^{4}\ \text{N/C}$ (given: elementary charge $e = 1.6\times10^{-19}\ \text{C}$; mass of the proton $m_p = 1.67\times10^{-27}\ \text{kg}$; take $g = 9.8\ \text{m/s}^2$).

1. a) Calculate the electric force on the proton.
2. b) Calculate its weight.
3. c) Calculate the ratio between both magnitudes and draw a conclusion: in particle physics, can gravity be neglected compared to the electric force?

Two point charges $q_1 = 4\,\mu\text{C}$ and $q_2 = -4\,\mu\text{C}$ occupy, respectively, the origin of coordinates and the point $(0, 3)$, with coordinates in meters (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). We want to find the field at point $P(4, 0)$. Observe the diagram:

1. a) Calculate the field vector $\vec{E}_1$ produced by $q_1$ at $P$.
2. b) Calculate the magnitude of the field produced by $q_2$ at $P$ and, taking into account that this field points from $P$ toward the negative charge (unit vector $-\frac{4}{5}\,\vec{i} + \frac{3}{5}\,\vec{j}$), write down the vector $\vec{E}_2$.
3. c) Apply the superposition principle: calculate the total field vector at $P$ and its magnitude.

Two point charges $q_1 = 1\,\mu\text{C}$ and $q_2 = 4\,\mu\text{C}$ lie on the $x$-axis, at $x_1 = 0$ and $x_2 = 3\ \text{m}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). The figure shows the three regions of the axis.

1. a) Reason, by analyzing the direction of the field from each charge, in which of the three regions of the axis (to the left of $q_1$, between the charges, or to the right of $q_2$) the total field can be zero.
2. b) Set up the condition for zero field in that region and calculate the exact point.
3. c) Check the result by verifying that at that point the two fields have the same magnitude.

Two point charges $q_1 = q$ and $q_2 = -2 \times q$, with $q > 0$, lie on the $x$-axis, at $x_1 = 0$ and $x_2 = 3\ \text{m}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Predict, justifying it using the direction of the fields and the absolute values of the charges, in which region of the axis the total field can be zero: between the charges or outside the segment? And on which side?
2. b) Set up the zero-field equation and solve it, obtaining the exact point.
3. c) Explain why the result does not depend on the value of $q$.

Three identical charges $q = 2\ \text{nC}$ occupy the vertices of an equilateral triangle of side length $30\ \text{cm}$ (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$). Observe the diagram:

1. a) Justify, by symmetry (the three field vectors at the center form angles of $120^\circ$ with each other), that the total field at the center of the triangle is zero.
2. b) One of the charges is removed. Taking into account that the field of all three was zero, reason that the field of the remaining two is exactly the opposite of the one produced by the removed charge, and calculate its magnitude (the distance from each vertex to the center is $\frac{0.3}{\sqrt{3}}\ \text{m}$).
3. c) State the direction and sense of the resultant field relative to the vertex that has been left empty.

Two identical charges $q = 5\,\mu\text{C}$ occupy two adjacent vertices, $A(0, 20)$ and $B(20, 20)$, of a square of side length $20\ \text{cm}$, with coordinates in centimeters (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$; $\cos 45^\circ \approx 0.71$). Vertex $D(0, 0)$ is studied. The figure shows the arrangement.

1. a) Calculate the field (magnitude, direction and sense) produced at $D$ by the charge located at $A$.
2. b) Calculate the magnitude of the field produced at $D$ by the charge located at $B$ (note that the distance is the diagonal of the square) and resolve it into its components.
3. c) Add the vectors and calculate the magnitude of the total field at $D$.
4. d) Calculate the magnitude of the force that a charge $Q = -2\,\mu\text{C}$ placed at $D$ would experience.

Your classmate calculates the electric field at the midpoint of the segment joining a charge $q_1 = 6\,\mu\text{C}$ to another charge $q_2 = -6\,\mu\text{C}$, separated by $20\ \text{cm}$, and writes: "the field is zero, because the charges are opposite and cancel out" (take $K = 9\times10^{9}\ \text{N}\times\text{m}^2/\text{C}^2$).

1. a) Explain the error: draw the vectors $\vec{E}_1$ and $\vec{E}_2$ at the midpoint in your notebook and reason about which direction each one points.
2. b) Calculate the correct field (magnitude and direction) at that point.
3. c) For which pair of charges with the same absolute value would your classmate's result be correct? Justify your answer.

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

In a chamber of a linear particle accelerator, there is a uniform vertical electric field $E = 5\times10^{3}\ \text{N/C}$. A proton and an electron enter it separately with the same horizontal velocity (given: elementary charge $e = 1.6\times10^{-19}\ \text{C}$; mass of the electron $m_e = 9.11\times10^{-31}\ \text{kg}$; mass of the proton $m_p = 1.67\times10^{-27}\ \text{kg}$).

1. a) Compare the magnitudes of the electric forces on both particles and explain why they are equal.
2. b) Calculate the acceleration of each particle.
3. c) Calculate the ratio $\frac{a_e}{a_p}$ and relate it to the mass ratio $\frac{m_p}{m_e}$.
4. d) Predict and justify: do both particles deflect to the same side? In the same time of flight, which one deflects more and by how many times?