The electric field strength at a point is defined as the force per unit charge exerted on a stationary positive charge at that point.
To define electric field strength, we imagine putting a positive test charge $+Q$ in the field and measuring the electric force F that it feels (Figure 21.10). It is important to recognise the importance of using a positive test charge, as this determines the direction of an electric field. (If you have used a charged gold leaf to investigate a field, this illustrates the principle of testing the field with a charge.)

From this definition, we can write an equation for E:

$E = \frac{F}{Q}$

where E is the electric field strength, F is the force on the charge and Q is the charge.
It follows that the units of electric field strength are newtons per coulomb ($N\,{C^{ - 1}}$).

The strength of a uniform field

You can set up a uniform field between two parallel metal plates by connecting them to the terminals of a high-voltage power supply (Figure 21.11). The strength of the field between them depends on two factors:
- the voltage V between the plates – the higher the voltage, the stronger the field: $E \propto V$
- the separation d between the plates – the greater their separation, the weaker the field: $E \propto \frac{1}{d}$
These factors can be combined to give an equation for E:

$E =  - \frac{V}{d}$

Worked example 1 shows a derivation of this. Note that the minus sign is often omitted because we are only interested in the magnitude of the field, not its direction. In Figure 21.11, the voltage V increases towards the right while the force F acts in the opposite direction, towards the left. E is a vector quantity.

If we look at this formula in a little more detail we can see that the electric field is really equal to the change in the potential (potential difference) divided by the change in distance (distance moved). This is written:

$E =  - \frac{{\Delta V}}{{\Delta d}}$

where the symbol $\Delta $ means ‘change of’.

From this equation, we can see that we can write the units of electric field strength as volts per metre ($V{m^{ - 1}}$). Note:

$1\,V{m^{ - 1}} = 1\,N{C^{ - 1}}$

Worked example 2 shows how to solve problems involving uniform fields.

Questions

 

4) Figure 21.12 shows an arrangement of parallel plates, each at a different voltage. The electric field lines are shown in the space between the first pair. Copy and complete the diagram to show the electric field lines in the other two spaces.

5) Calculate the electric field strength at a point where a charge of $20 mC$ experiences a force vertically downwards of $150 N$.

6) Calculate the electric field strength between two parallel charged plates, separated by $40 cm$ and with a potential difference between them of $1000 V$.

7) An electron is situated in a uniform electric field. The electric force that acts on it is $8 \times {10^{ - 16}}\,N$.
What is the strength of the electric field? (Electron charge $e = 1.6 \times {10^{ - 19}}C$.)

8) Air is usually a good insulator. However, a spark can jump through dry air when the electric field strength is greater than about $40000V\,c{m^{ - 1}}$. This is called electrical breakdown. The spark shows that electrical charge is passing through the air–there is a current. (Do not confuse this with a chemical spark such as you might see when watching fireworks; in that case, small particles of a chemical substance are burning quickly.)
a: A Van de Graaff generator (Figure 21.13) is able to make sparks jump across a $4 cm$ gap. Estimate the voltage produced by the generator?
b: The highest voltage reached by the live wire of a conventional mains supply is $325 V$. In theory (but DO NOT try this), how close would you have to get to a live wire to get a shock from it?
c: Estimate the voltage of a thundercloud from which lightning strikes the ground $100 m$ below.