Electric Field and Applications of Electrostatics
Senior 2 · Physics
Electric Field and Applications of Electrostatics
Lesson 3 of 5
22 min
Your phone charger is labelled 5 V, a car battery 12 V, and the mains supply in Rwanda about 230 V. The base of a thundercloud can be at a potential of many millions of volts compared with the ground. The volt measures electric potential, a way of describing a field by energy rather than by force. Energy is often the easier route to an answer, just as it was for falling mangoes.
When you lift a stone, you do work against gravity and the stone gains gravitational potential energy. In the same way, when you push a positive charge towards another positive charge, you do work against the electric repulsion, and the charge gains electric potential energy. Let go, and that energy turns into kinetic energy as the charge is pushed away.
The electric potential V at a point is the work done per unit positive charge in bringing a charge from infinitely far away to that point:
V = W / q
Its unit is the volt (V): 1 V = 1 joule per coulomb (J/C). Unlike the field, potential is a scalar: it has a size and a sign but no direction.
Moving a charge of 3.0 μC between two points whose potentials differ by 12 V takes work
W = qV = 3.0 × 10⁻⁶ × 12 = 3.6 × 10⁻⁵ J = 36 μJ.
The potential difference (p.d.) between two points is the work done per unit charge in moving a charge between them. This is exactly what a voltmeter measures, and why the p.d. of a cell is its "voltage". An electron moved through 1 V gains 1.6 × 10⁻¹⁹ J; physicists call this amount of energy one electron-volt.
For a point charge Q, the potential at distance r is
V = kQ / r (k = 9.0 × 10⁹ N·m²/C²)
Note: r, not r². V is positive near a positive charge and negative near a negative charge.
Worked example. A charge of +2.0 nC sits at a point.
Because potential is a scalar, the potential due to several charges is just the sum of their potentials, with signs. At the midpoint between +4.0 nC and −4.0 nC, the two potentials are equal and opposite, so V = 0. But the field there is not zero: both fields point towards the negative charge and add. Zero potential does not mean zero field.
Between two parallel plates with a p.d. V across a gap d, the field is uniform and
E = V / d
This is why the field can be given in volts per metre (V/m), which is the same as N/C.
Worked example. Two plates 2.0 cm apart are connected to a 400 V supply. E = 400 ÷ 0.020 = 20 000 V/m (2.0 × 10⁴ N/C).
A line or surface on which every point has the same potential is an equipotential. No work is done moving a charge along an equipotential, so equipotentials always cross field lines at right angles. Around a point charge they are circles (spheres, in three dimensions); between parallel plates they are planes parallel to the plates.
In Charges and Fields, place a positive charge and use the voltage tool to read the potential at different distances: halve the distance and the potential doubles. Use the tool to draw equipotential lines, and check that they cross the field arrows at right angles.
Tip
Field or potential? Check the power of r. Field: E = kQ/r² (vector, direction matters). Potential: V = kQ/r (scalar, just add with signs).
Two formulas you will use again and again: W = qV for work or energy, and E = V/d for parallel plates (d in metres!).
Take k = 9.0 × 10⁹ N·m²/C².
How much energy does a charge of 2.0 C gain when it moves through a p.d. of 230 V?
Solution: W = qV = 2.0 × 230 = 460 J.
Find the potential 0.50 m from a charge of −3.0 nC.
Solution: V = kQ/r = 9.0 × 10⁹ × ( − 3.0 × 10⁻⁹) ÷ 0.50 = −54 V.
Two plates 5.0 mm apart have a field of 3.0 × 10⁴ V/m between them. What is the p.d.?
Solution: V = Ed = 3.0 × 10⁴ × 0.0050 = 150 V.
Charges of +2.0 nC and +2.0 nC are 0.20 m apart. Find the potential at the midpoint.
Solution: Each gives V = 9.0 × 10⁹ × 2.0 × 10⁻⁹ ÷ 0.10 = 180 V. Potentials add as scalars: 360 V. (The field there, by contrast, is zero: this is a neutral point.)
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