Electric Field and Applications of Electrostatics
Senior 2 · Physics
Electric Field and Applications of Electrostatics
Lesson 1 of 5
20 min
Comb your dry hair and the comb pulls small pieces of paper towards it without touching them. On a stormy afternoon in the rainy season, a thundercloud several kilometres high makes the hairs on your arm stand up before any lightning strikes. Charged objects push and pull on other charges at a distance. The idea that explains how is the electric field, and it is the key to lightning conductors, photocopiers and every electronic device you own.
You already know that there are two kinds of electric charge, positive and negative, that like charges repel and unlike charges attract, and that charge is measured in coulombs (C). Everyday charges are tiny, so you will meet the microcoulomb (1 μC = 10⁻⁶ C) and the nanocoulomb (1 nC = 10⁻⁹ C).
Instead of saying "charge A pushes charge B", physicists say: charge A creates an electric field in the space around it, and the field pushes on any charge placed in it.
An electric field is a region in which an electric charge experiences an electric force.
To measure a field at a point, place a small test charge there and measure the force on it. Suppose a test charge of +2.0 μC placed at point P feels a force of 6.0 × 10⁻³ N towards the east.
Force per coulomb = 6.0 × 10⁻³ N ÷ 2.0 × 10⁻⁶ C = 3000 N/C, towards the east.
A test charge twice as big would feel twice the force, but the force per coulomb would be the same. That quantity describes the field itself.
The electric field intensity (or electric field strength) E at a point is the force per unit positive charge placed at that point:
E = F / q
Worked example (continued). At the same point P, replace the test charge with a charge of −4.0 μC. The force is F = qE = 4.0 × 10⁻⁶ × 3000 = 0.012 N, towards the west, because the charge is negative.
For a single point charge Q, the field at distance r is
E = kQ / r², where k = 9.0 × 10⁹ N·m²/C².
Worked example. Find the field 0.10 m from a charge of +5.0 nC.
E = 9.0 × 10⁹ × 5.0 × 10⁻⁹ ÷ (0.10)² = 45 ÷ 0.010 = 4500 N/C, pointing directly away from the charge.
At 0.20 m (twice as far): E = 45 ÷ 0.040 = 1125 N/C, a quarter as much. The field obeys an inverse-square law: double the distance, quarter the field.
Worked example. An electron (charge −1.6 × 10⁻¹⁹ C) is in a field of 2.0 × 10⁴ N/C. The force on it is F = qE = 1.6 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N, opposite to the field.
Open Charges and Fields. Drag one positive charge onto the screen and tick Electric Field. Then drag a yellow E-field sensor near the charge and slowly away from it. Watch the arrow shrink and read the value: move twice as far and the field drops to about a quarter. Swap in a negative charge and watch the arrows turn round to point inwards.
Tip
Units first, every time. Convert μC and nC to coulombs, and cm to metres, before you substitute. Then square r: forgetting to square the distance is the most common error in E = kQ/r².
Direction: "away from +, towards −". Always state the direction with the size; a field value without a direction is only half an answer.
Take k = 9.0 × 10⁹ N·m²/C².
A charge of +5.0 μC feels a force of 0.020 N at a point. Find the field intensity there.
Solution: E = F/q = 0.020 ÷ 5.0 × 10⁻⁶ = 4000 N/C, in the direction of the force.
In a field of 500 N/C pointing north, what force acts on a charge of −3.0 μC?
Solution: F = qE = 3.0 × 10⁻⁶ × 500 = 1.5 × 10⁻³ N, towards the south (opposite to the field, because the charge is negative).
Find the field 0.30 m from a point charge of +2.0 μC.
Solution: E = 9.0 × 10⁹ × 2.0 × 10⁻⁶ ÷ 0.30² = 18 000 ÷ 0.090 = 2.0 × 10⁵ N/C, away from the charge.
The field 5 cm from a small charged sphere is 800 N/C. What is it 10 cm from the sphere?
Solution: Doubling the distance divides the field by 2² = 4: 800 ÷ 4 = 200 N/C.
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