Electric fields and field lines
Section: Electric Fields | Syllabus: Cambridge A Level Physics (9702)
Origins of Static Electricity
About 2600 years ago, Thales of Miletus discovered that rubbing amber with fur allowed it to attract feathers. This is charging by friction. Today, we observe this with plastic rulers attracting paper or balloons bending water.
Real-World Examples- Plastic Ruler & Paper: Rubbing transfers electrons to the ruler. When brought near paper, it causes a slight redistribution of charge (induction), making the side closest to the ruler oppositely charged, resulting in attraction.
- Balloon & Water: A charged balloon attracts a stream of water because water molecules are permanent dipoles. The molecules align their opposite poles toward the balloon, creating an attractive force.
Defining the Electric Field
An electric field is a region around a charged particle where an electrical force is exerted on other charges. It is represented by field lines.
Field DirectionThe direction of an electric field at a point is the direction of the force on a small positive test charge placed at that point.
Note: The test charge is "small" so its own field doesn't significantly distort the field being measured.
E = (F/q)Where E is electric field strength (N C^-1), F is force (N), and q is charge (C).
Electric Field Patterns
Field lines show the path a positive test charge would take. They always point out of positive and into negative charges.
Figure 18.3: Common Patterns Isolated Sphere: Radial lines (outwards for +, inwards for -).
Opposite Charges: Curved lines connecting + to -.
Like Charges: Lines curve away from each other, showing repulsion.
Worked Examples
Worked Example 1: Force on an IonQuestion: Calculate the force on a positively charged ion (charge 1.6 × 10^-19 C) drifting into a field of strength 2.0 × 10^5 N C^-1.
SolutionF = Eq = (2.0 × 10^5) × (1.6 × 10^-19) = 3.2 × 10^-14 N.
Worked Example 2: Field Strength CalculationQuestion: A charge of 1.5~ experiences a force of 1.2 × 10^-3 N. Calculate the field strength.
SolutionE = F / q = (1.2 × 10^-3) / (1.5 × 10^-6) = 800 N C^-1.
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