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 Direction The 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 Ion Question: 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.
Solution F = Eq = (2.0 × 10^5) × (1.6 × 10^-19) = 3.2 × 10^-14 N. Worked Example 2: Field Strength Calculation Question: A charge of 1.5~μC experiences a force of 1.2 × 10^-3 N. Calculate the field strength.
Solution E = F / q = (1.2 × 10^-3) / (1.5 × 10^-6) = 800 N C^-1.
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