Coulomb law and electric-field superposition

Coulomb's law gives the force between two point charges in vacuum:

$$F=k\,\frac{|q_1q_2|}{r^{2}},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx 9.0\cdot10^{9}\ \text{N}\cdot\text{m}^{2}/\text{C}^{2}.$$

The force acts along the line joining the charges: like charges repel, opposite charges attract, and by Newton's third law the forces on $q_1$ and $q_2$ are equal in magnitude and opposite in direction.

The field of a stationary point charge at distance $r$ is

$$\vec{E}=k\,\frac{q}{r^{2}}\,\hat{r},\qquad \vec{F}=q_0\vec{E},$$

where $\hat{r}$ is the unit vector from the charge to the observation point.

The superposition principle states that the field of a system of charges is the vector sum of the fields of the individual charges: $\vec{E}=\sum_i \vec{E}_i$; because the addition is vectorial, it is convenient to resolve the fields into components, $\vec{E}=E_x\hat{x}+E_y\hat{y}$.

Finally, any field calculation rests on Newton's laws: once $\vec{E}$ is known, the force on a test charge $q_0$ is $\vec{F}=q_0\vec{E}$.

Two point charges $q_1=+2.0\ \mu\text{C}$ and $q_2=-3.0\ \mu\text{C}$ are separated by a distance $r=0.30\ \text{m}$. Find the magnitude and direction of the force on $q_2$ and the force on $q_1$. Take $k=9.0\cdot10^{9}\ \text{N}\cdot\text{m}^{2}/\text{C}^{2}$.

Point charges $q_1=+5.0\ \text{nC}$ and $q_2=+5.0\ \text{nC}$ are fixed at $(0,\,0)$ and $(0.60,\,0)$ m. Find the electric-field vector and its magnitude at point $P(0.30,\,0.40)$ m.

Charges $q_1=+9.0\ \text{nC}$ and $q_2=+4.0\ \text{nC}$ are fixed at $x=0$ and $x=0.50\ \text{m}$ on the $Ox$ axis. Find the coordinate of the point where the resultant field is zero. Are there any other such points?

Safe self-check in a simulation (no real high voltages and no handling of charged capacitors).

1. Open an interactive electric-field model (for example, PhET "Charges and Fields"). Place $q_1=+5.0\ \text{nC}$ at the origin and $q_2=+5.0\ \text{nC}$ at $(0.60,\,0)$ m.
2. Use the field sensor to measure $\vec{E}$ at $(0.30,\,0.40)$ m and compare it with the calculated $(0,\ 288)\ \text{N/C}$; if needed, switch the sensor to show the components $E_x$ and $E_y$.
3. Move the sensor along the $Ox$ axis in steps of $0.05$ m and record $E_x$; check that at $x=0.30$ m the field becomes zero (equal like charges cancel at the midpoint of the segment).
4. Replace $q_2$ with $-5.0\ \text{nC}$ and repeat the measurements: there is no zero-field point between the charges any more, because the fields point the same way; explain this using the directions of the unit vectors $\hat{r}_i$.
5. Sum up: write for the control point the equality $\vec{E}=\sum_i k q_i\hat{r}_i/r_i^{2}$ with the numbers substituted and compare it with the sensor readings.

Show that the resultant field at a point is the vector sum of the fields of the individual charges, and that the zero-field point lies between like charges.

Diagram: two positive charges $q_1$ and $q_2$ on the $Ox$ axis; at point $P$ the arrows $\vec{E}_1$ and $\vec{E}_2$ are drawn along the unit vectors $\hat{r}_1$ and $\hat{r}_2$, with the parallelogram of their sum $\vec{E}$. Beside it, a graph of $E_x(x)$ along the $Ox$ axis with a zero at $x=0.30$ m.

After rubbing plastic wrap with a wool cloth, it receives charge $q_1 = +8\,\mathrm{nC}$; a pepper flake has charge $q_2 = -2\,\mathrm{nC}$ at distance $r = 1.5\,\mathrm{cm}=0.015\,\mathrm{m}$. The Coulomb force magnitude is $F = k\frac{|q_1 q_2|}{r^2} = 9\times10^9 \frac{8\times10^{-9}\cdot 2\times10^{-9}}{(0.015)^2} \approx 6.4\times10^{-4}\,\mathrm{N}.$ The weight of a flake of mass $m=0.05\,\mathrm{g}=5\times10^{-5}\,\mathrm{kg}$ is $mg \approx 4.9\times10^{-4}\,\mathrm{N}$, so the electric attraction is comparable to its weight and can lift it.

The scenario uses Coulomb's law: the force is inversely proportional to $r^2$; if the distance increases to $3\,\mathrm{cm}$, the force drops to $1.6\times10^{-4}\,\mathrm{N}$.

A cloud contains two compact charged centres: $Q_1 = -10\,\mathrm{C}$ at height $h_1=600\,\mathrm{m}$ and $Q_2 = -15\,\mathrm{C}$ at height $h_2=900\,\mathrm{m}$ above one point on the ground. Their fields at this point are $E_1 = k\frac{|Q_1|}{h_1^2} = 9\times10^9\frac{10}{600^2} = 2.5\times10^5\,\mathrm{V/m},$ $E_2 = k\frac{|Q_2|}{h_2^2} = 9\times10^9\frac{15}{900^2} \approx 1.67\times10^5\,\mathrm{V/m}.$ Both vectors point upward, so superposition gives $E = E_1+E_2 \approx 4.17\times10^5\,\mathrm{V/m}.$ This is below the breakdown value $3\times10^6\,\mathrm{V/m}$, but additional charged centres can add their fields up to breakdown.

The scenario exercises vector superposition of fields from point charges: for parallel vectors the magnitudes add, while for non-parallel vectors one must add them by the parallelogram rule.

On dry artificial turf, player A receives charge $q_A=+4\,\mu\mathrm{C}$, player B receives $q_B=+6\,\mu\mathrm{C}$ at distance $r_{AB}=1.5\,\mathrm{m}$, and player C receives $q_C=+2\,\mu\mathrm{C}$ exactly midway between A and B. The force on C from A is $F_{AC}=k\frac{q_A q_C}{(0.75)^2}=9\times10^9\frac{4\times10^{-6}\cdot 2\times10^{-6}}{0.5625}\approx 0.128\,\mathrm{N}$ directed to the right. From B: $F_{BC}=k\frac{q_B q_C}{(0.75)^2}=9\times10^9\frac{6\times10^{-6}\cdot 2\times10^{-6}}{0.5625}\approx 0.192\,\mathrm{N}$ directed to the left. The net force on C is $F_{\text{net}} = 0.192 - 0.128 = 0.064\,\mathrm{N}$ to the left.

Here Coulomb-force superposition applies: each pair of charges gives its own force, and the resultant is found by vector addition. If the mass of C is $m=70\,\mathrm{kg}$, Newton's second law gives $a=F_{\text{net}}/m \approx 9.1\times10^{-4}\,\mathrm{m/s^2}$.