Vector treatment of the electric field

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 $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$). 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.

A space probe accumulates a net charge of $2\ \text{nC}$ due to bombardment by solar particles (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$). 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 metres).
2. b) Also calculate the field at $1\ \text{m}$ and check by what factor it has decreased when multiplying the distance by ten.

A sensor measures an electric field of $4500\ \text{N/C}$ at $2\ \text{m}$ from the centre of a charged metal sphere in the laboratory; the field points towards the sphere (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$). The situation is represented in the figure.

1. a) Rearrange for charge in the expression $E = K \cdot \frac{Q}{r^2}$ 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.

From the definition $E = \frac{F}{q}$, state which of these expressions are valid units for electric field strength and reason each answer:

1. a) $\text{N/C}$
2. b) $\text{N}\cdot\text{C}$
3. c) $\text{V/m}$ (note that $1\ \text{V} = 1\ \text{J/C}$)
4. d) $\text{J/C}$

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) Reason why, since the charge is negative, the force is in the opposite direction to the field.

In a cathode ray tube, an electron passes through a region with a uniform electric field $E = 1000\ \text{N/C}$ (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; mass of the electron $m_e = 9,11\cdot10^{-31}\ \text{kg}$).

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

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 $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$). 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.

A charge $Q = 8\ \text{nC}$ is fixed to an insulating stand in the laboratory (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

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

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 metres (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$). 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 $P$ towards 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.

In an inkjet printer, charged ink droplets are deflected by an electric field. To study the principle, consider an electron entering with horizontal velocity $v_0 = 3\cdot10^{7}\ \text{m/s}$ into a region of length $L = 10\ \text{cm}$ where there is a uniform vertical field $E = 10^{4}\ \text{N/C}$, perpendicular to its velocity (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; mass of the electron $m_e = 9,11\cdot10^{-31}\ \text{kg}$). The diagram shows the setup.

1. a) Calculate the acceleration of the electron in the field region and state its direction and sense.
2. b) Calculate the time it takes to cross the region.
3. c) Assuming horizontal motion with constant velocity and vertical motion with constant acceleration (like in projectile motion), calculate the vertical deflection at the exit, $y = \frac{1}{2} \cdot a \cdot t^2$.
4. d) Reason what shape the trajectory has inside the field and what type of motion the electron executes after leaving it.

Two point charges $q_1 = q$ and $q_2 = -2 \cdot q$, with $q > 0$, lie on the $x$-axis, at $x_1 = 0$ and $x_2 = 3\ \text{m}$ (take $K = 9\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$).

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 $q$.

In a chamber of a linear particle accelerator, there is a uniform vertical electric field $E = 5\cdot10^{3}\ \text{N/C}$. A proton and an electron enter it separately with the same horizontal velocity (given: elementary charge $e = 1,6\cdot10^{-19}\ \text{C}$; mass of the electron $m_e = 9,11\cdot10^{-31}\ \text{kg}$; mass of the proton $m_p = 1,67\cdot10^{-27}\ \text{kg}$).

1. a) Compare the magnitudes of the electric forces on both particles and reason 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?