Electric charge, Coulomb's law and the field of point charges

The first week of Physics II sets up the language for the rest of the semester. Gauss’s law, potential, capacitors, all of them rest on the two formulas you learn here: the Coulomb force and the electric field of a point charge. The question type that comes from this week is almost always the same: a few point charges are given, and you are asked for the net force or field at some point. Students lose points not because they don’t know the physics, but because they add vectors like scalars.

Golden rules:

1. Find the magnitude with absolute values, find the direction from the picture. Don’t plug signed charges into the formula and try to read the direction from the sign.
2. Convert units to SI at the start. μC → $10^{-6}$ C, nC → $10^{-9}$ C, cm → m. Then square.
3. Split vectors into components. Add the $x$‘s among themselves and the $y$‘s among themselves.
4. Look for symmetry. Fields that cancel save you half a page of calculation.

Electric charge

There are two kinds of charge in nature: positive and negative. Like charges repel, opposite charges attract. An object is normally neutral: it has equal numbers of protons and electrons. “Charging” an object means adding or removing electrons. Protons are fixed in the nucleus; it is always the electrons that move.

Three properties of charge you will use on the exam:

The elementary charge is $e = 1.60 \times 10^{-19}$ C. The electron’s charge is $-e$, the proton’s is $+e$. The SI unit is the coulomb (C), but the charges you will see on the exam are usually of the order of microcoulombs ($\mu$C) or nanocoulombs (nC).

Example 1. An object has a charge of $-8.0$ nC. How many excess electrons does it have?

$N = \frac{|q|}{e} = \frac{8.0 \times 10^{-9}\ \text{C}}{1.60 \times 10^{-19}\ \text{C}} = 5.0 \times 10^{10}$

Negative charge means excess electrons. If it were positive, the same number of electrons would be missing.

Conductors and insulators

Conductor: It contains electrons that can move freely. Metals, the human body, salt water. If you put charge on a conductor, it quickly spreads over the whole surface.

Insulator: The electrons are tightly bound to atoms. Glass, plastic, rubber, dry wood. If you charge an insulator by rubbing, the charge stays where you put it.

Semiconductor: Silicon, germanium. Its conductivity lies between the two and can be adjusted by doping.

Charging by induction

Bring a negatively charged rod near a neutral metal sphere without touching it. The sphere’s electrons are repelled and move to the far side. The near side becomes positive, the far side negative. Now connect the far side to ground with a wire: electrons flow into the ground. First remove the wire, then take the rod away. The sphere is now positively charged, and the rod has lost no charge at all.

PITFALL · Order matters. If you take the rod away before removing the ground connection, electrons flow back from the ground and the sphere stays neutral. That is why the exam asks “in what order” questions.

In insulators a similar effect is weaker: the charge distribution of the molecules shifts slightly (polarization). This is why a charged comb attracts small pieces of paper.

Coulomb’s law

The magnitude of the force between two point charges is:

$F_e = k_e \frac{|q_1|\,|q_2|}{r^2}, \qquad k_e = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N·m}^2/\text{C}^2$

Here $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N·m}^2)$ is the permittivity of free space. The force acts along the line joining the two charges. Newton’s third law holds: the force $q_1$ exerts on $q_2$ is equal and opposite to the force $q_2$ exerts on $q_1$, even when the charges have different magnitudes.

Vector form:

$\vec F_{12} = k_e \frac{q_1 q_2}{r^2}\,\hat r_{12}$

$\hat r_{12}$ is the unit vector pointing from $q_1$ to $q_2$. If $q_1 q_2 > 0$, the force pushes $q_2$ away from $q_1$.

Example 2. Charges of $+3.0\ \mu$C and $-5.0\ \mu$C are 0.20 m apart.

$F = \frac{(8.99 \times 10^{9})(3.0 \times 10^{-6})(5.0 \times 10^{-6})}{(0.20)^2} = 3.37\ \text{N}\ \ (\text{attraction})$

Several charges: superposition

The net force on a charge is the vector sum of the forces that each of the other charges alone exerts on it. The method is always the same:

1. For each pair, find the magnitude using absolute values.
2. Draw the direction: repulsion or attraction?
3. Split each force into $x$ and $y$ components.
4. Add the components, then find the magnitude and angle.

Example 3. $q_1 = +2.0\ \mu$C at the origin, $q_2 = -3.0\ \mu$C at $(0.30, 0)$ m, $q_3 = +4.0\ \mu$C at $(0, 0.40)$ m. What is the net force on $q_3$?

q1 → q3:  r = 0.40 m, repulsion (pushes q3 away from q1, +y)
  F1 = (8.99e9)(2.0e-6)(4.0e-6) / 0.40² = 0.449 N
  F1 = (0, +0.449) N

q2 → q3:  r = √(0.30² + 0.40²) = 0.50 m, attraction (from q3 toward q2)
  F2 = (8.99e9)(3.0e-6)(4.0e-6) / 0.50² = 0.432 N
  direction from q3 to q2: (0.30, -0.40)/0.50 = (0.60, -0.80)
  F2 = (0.432·0.60, -0.432·0.80) = (+0.259, -0.345) N

Total:  Fx = 0.259 N,  Fy = 0.449 - 0.345 = 0.104 N
  |F| = √(0.259² + 0.104²) = 0.279 N
  θ = arctan(0.104 / 0.259) = 21.9°  (from the +x axis, first quadrant)

TIP · 3-4-5 triangles. In exam questions the distances are often in a 3-4-5 ratio. You can immediately write the cosine and sine as 0.60 and 0.80.

Electric field

A charge changes the space around it. Any charge placed there feels this change as a force. This change is called the electric field. It is defined as the force on a small positive test charge $q_0$, divided by that charge:

$\vec E = \frac{\vec F_e}{q_0} \qquad [\text{N/C}]$

Once the field is known, the force on any charge $q$ placed there is:

$\vec F_e = q\vec E$

If $q$ is positive the force points along the field, if it is negative the force points opposite to it. This is the first thing to check in questions about the direction an electron moves.

The field of a point charge:

$E = k_e \frac{|q|}{r^2}$

Direction: away from a positive charge (radially outward), toward a negative charge (inward).

Superposition for the field of point charges

Just as with force: $\vec E = \sum_i \vec E_i$. Draw each charge’s field separately, split into components, and add.

Example 4. $+4.0$ nC at $(-0.10, 0)$ m, $-4.0$ nC at $(+0.10, 0)$ m. What is the field at the point $P = (0, 0.10)$ m?

Distance to both charges:  r = √(0.10² + 0.10²) = 0.1414 m,  r² = 0.020 m²
Magnitude (same for both):  E = (8.99e9)(4.0e-9) / 0.020 = 1798 N/C

E1 (+ charge, AWAY from it): direction (0.10, 0.10)/0.1414 = (0.707, 0.707)
  E1 = (+1271, +1271) N/C
E2 (− charge, TOWARD it): direction from P to (0.10, 0) is (0.10, -0.10)/0.1414 = (0.707, -0.707)
  E2 = (+1271, -1271) N/C

Total: Ex = 2543 N/C, Ey = 0
  E = 2.54 × 10³ N/C, in the +x direction

The $y$ components cancelled because of symmetry. This shows that on the perpendicular bisector of two equal and opposite charges (a dipole), the field is always parallel to the line from the positive to the negative charge.

The point where the field is zero

Finding where the fields of two point charges cancel is a classic question. First think about the region, then set up the equation:

When setting up the equation, write $\dfrac{k|q_1|}{r_1^2} = \dfrac{k|q_2|}{r_2^2}$, then take the square root of both sides. You don’t need to solve a quadratic.

PITFALL · Don’t look for a zero in between. Between two opposite charges both fields point the same way (from the positive toward the negative). If you found a zero point there, you picked the wrong region.

Common mistakes