describe electric fields using the field model, apply Coulomb's law $F = k q_1 q_2 / r^2$ and the relationships $E = F/q$, $E = kQ/r^2$ for point charges and $E = V/d$ for the uniform field between parallel plates; identify the directions of field, force and acceleration of charged particles in uniform and radial fields

What this dot point is asking

VCAA wants you to describe the electric field using the field model, apply Coulomb's law for point charges, calculate field strength for a point charge ($E = kQ/r^2$) and between parallel plates ($E = V/d$), and predict the direction of force and acceleration on a charged particle.

The answer

The field model

An electric field surrounds every electric charge. Another charge placed in the field experiences a force. The field is drawn with field lines that point in the direction of the force on a positive test charge.

• Field lines start on positive charges and end on negative charges (or extend to infinity).
• The density of field lines indicates field strength.
• Between two parallel oppositely charged plates, field lines run from positive to negative, parallel and evenly spaced (uniform field).

Coulomb's law

The electric force between two point charges $q_1$ and $q_2$ separated by distance $r$ is:

where $k = 9.0 \times 10^9$ N m squared per C squared. Like signs repel; opposite signs attract.

Electric field strength

The electric field strength $E$ at a point is the force per unit positive test charge:

Units are N/C, equivalent to V/m.

Point charge. A single point charge $Q$ produces a radial field:

pointing outward from a positive charge, inward toward a negative charge.

Parallel plates. Two large, parallel, oppositely charged plates separated by distance $d$ with potential difference $V$ produce a uniform field in the gap:

The field points from the positive plate to the negative plate.

Force and acceleration on a charged particle

A charge $q$ in field $E$ feels force $F = qE$. By Newton's second law its acceleration is:

Direction. A positive charge accelerates along the field. A negative charge accelerates opposite to the field.

In a uniform field between parallel plates, a charged particle entering perpendicular to the field follows a parabolic path (analogous to projectile motion with $g$ replaced by $a = qE/m$).

Comparing electric and gravitational fields

The two laws have the same $1/r^2$ form. The key difference is that gravity is always attractive, while the electric force can be either attractive or repulsive.

Worked example with numbers

An electron is released from rest near the negative plate of a parallel-plate setup with $V = 200$ V and $d = 4.0$ mm. Find the acceleration of the electron and the speed it reaches at the positive plate. ($m_e = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C.)

Field: $E = V/d = 200 / 0.004 = 5.0 \times 10^4$ V/m.

Force: $F = eE = 1.6 \times 10^{-19} \times 5.0 \times 10^4 = 8.0 \times 10^{-15}$ N (toward the positive plate).

Acceleration: $a = F/m = 8.0 \times 10^{-15} / 9.1 \times 10^{-31} = 8.8 \times 10^{15}$ m/s squared.

Speed using $v^2 = 2 a d$: $v = \sqrt{2 \times 8.8 \times 10^{15} \times 0.004} = \sqrt{7.0 \times 10^{13}} = 8.4 \times 10^6$ m/s.

(The same answer comes from energy: $\frac{1}{2} m v^2 = eV$, $v = \sqrt{2eV/m}$.)

Try it: Electric field calculator - enter charge and distance, or voltage and plate separation, and get $E$ and $F$.

Common traps

Confusing the direction of force on a negative charge. The field points from positive to negative; the force on a negative charge is opposite to the field, so it accelerates toward the positive plate.

Mixing units of $E$. N/C and V/m are the same unit. Use whichever matches the rest of the question.

Forgetting to square the radius. $E = kQ/r^2$, not $kQ/r$.

Applying $E = V/d$ for a point charge. $E = V/d$ only works for a uniform field (between parallel plates or inside a long parallel-plate capacitor). Use $E = kQ/r^2$ for a point charge.

Using nano and milli without converting. $1$ nC $= 10^{-9}$ C; $1$ mm $= 10^{-3}$ m. Convert before substituting.

In one sentence

An electric field has strength $E = kQ/r^2$ around a point charge and $E = V/d$ in the uniform region between parallel plates; a charge $q$ in the field feels force $F = qE$ that is along the field for positive charges and opposite for negative charges.