The Memory Trick
💡 E = F/q — Force PER Unit Charge
The electric field at any point in space is defined as the force a positive test charge would experience there, divided by the magnitude of that test charge: E = F/q. This division is what makes the electric field a property of the space itself (created by other charges), independent of whatever specific test charge you might place there to measure it.
Why It Works
Dividing out the test charge's magnitude means the electric field describes 'what any charge placed here would feel, per unit of its own charge' — a small test charge and a large one at the same location would feel proportionally different total forces, but they'd calculate the exact same electric field value, because the field itself doesn't depend on the charge used to measure it.
Step by Step
Working With Electric Fields
1
Field lines run from positive to negative
Electric field lines originate at positive charges and terminate at negative charges, showing the direction a positive test charge would be pushed at each point.
Between a positive and negative charge, field lines curve directly from the positive charge to the negative charge.
2
Line density indicates field strength
Where field lines are drawn closer together, the electric field is stronger; where they're spread farther apart, the field is weaker.
Near a point charge, field lines are densely packed close to the charge and spread out (weaken) with increasing distance — visually representing the inverse-square falloff.
3
Uniform field between parallel plates
A special, commonly-tested case: the field between two oppositely-charged parallel plates is uniform (constant magnitude and direction) and given simply by E = V/d, where V is the voltage between the plates and d is their separation.
This uniform-field configuration is the basis of parallel-plate capacitors and many charged-particle acceleration devices.
🏥 Worked Example
A parallel plate capacitor has plates separated by 0.02 m with a voltage of 100 V across them. What is the electric field between the plates, and what force would a +5×10⁻⁶ C charge experience there?
1
Find the field using E = V/d: E = 100/0.02 = 5,000 V/m (or N/C).
2
Find the force using F = Eq (rearranged from E = F/q): F = 5,000 × 5×10⁻⁶.
3
Solve: F = 0.025 N — the force this specific charge would experience, calculated directly from the field value that describes the space between the plates.
📌 Exam Application
Exams test correctly applying E = F/q (and its rearrangement F = Eq), calculating uniform field strength between parallel plates using E = V/d, and correctly interpreting field line diagrams for direction and relative strength.
⚠️ Most Common Electric Fields Mistakes
The most common trap is confusing electric FIELD (E, a property of space, in N/C or V/m) with electric FORCE (F, what a specific charge experiences, in Newtons) — the field exists at a point in space regardless of whether any charge is actually there to feel a force from it.
✓ Quick Self-Test
1) Write the definition of electric field in terms of force and charge. E = F/q. 2) In which direction do electric field lines point — from negative to positive, or positive to negative? From positive to negative. 3) What does a denser concentration of field lines in a diagram indicate? A stronger electric field at that location. 4) Write the formula for the electric field between two parallel plates. E = V/d. 5) Why is electric field considered a property of space itself, rather than of any specific charge placed there? Because dividing out the test charge (E = F/q) removes dependence on the specific test charge's magnitude — the same field value applies regardless of what test charge measures it.
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