Potential difference
When a charged object is moved between two points in an electric field, work is done. Work is a measure of the energy that has been transferred and is measured in joules (J).
The charge of an object is measured in coulombs (C).
If the object has a charge of one coulomb, and the work done in moving the charge between the two points is one joule the potential difference between the points is one volt.
This allows us to use the equation \(W = QV\) where:
- \(W\) is the work done
- \(Q\) is the charge on the object
- \(V\) is the potential difference
Sometimes this equation is written as \(E_{W}=QV\) where \(E\) is the energy.
Question
When 5 C of charge are moved between two points in an electric field 20 J of work are done. Calculate the potential difference between the two points.
\({E_W} = 20J\)
\(Q = 5C\)
\(V = \frac{{{E_W}}}{Q}\)
\(V = \frac{{20}}{5}\)
\(V = 4V\)
The potential difference between the points is 4 V.
The energy transferred to a charged particle will cause it to gain kinetic energy, increasing its velocity.
If we know how much work is done on the charge as it passes through the electrical field, we can find its velocity.
Work done, \(W=QV\) = kinetic energy gained, \({E_K} = \frac{1}{2}m{v^2}\)
Question
An electron has a charge of \(1.6 \times 10^{-19}C\), and a mass of \(9.11 \times 10^{-31}kg\). It is held stationary in an electric field. The particle is released and allowed to accelerate through a potential difference of \(2 kV\).
Calculate the energy gained by the electron due to the field.
\(Q=1.6 \times 10^{-19}C\)
\(V=2000V\)
\(E=QV\)
\(E=1.6 \times 10^{-19}\times 2000\)
\(E= 3.2 \times 10^{-16}J\)
This is equivalent to the amount of kinetic energy gained by the electron.
Question
Now calculate the final speed.
\(E_{K}=3.2 \times 10^{-16}J\)
\(m=9.11 \times 10^{-31}kg\)
\({E_K} = \frac{1}{2}m{v^2}\)
\(v= \sqrt {\frac{2E_{K}}{m}}\)
\(v= \sqrt {\frac{2 \times 3.2 \times 10^{-16}}{9.11 \times 10^{-31}}}\)
\(v= 2.7 \times 10^{7}ms^{-1}\)
The final speed of the electron is \(2.7 \times 10^{7}ms^{-1}\)
Question
If an electron is accelerated by a potential difference four times greater than the original, will the speed be four times greater?
No, the kinetic energy will be four times greater but the speed is squared to give kinetic energy so the speed will only be doubled.
This is one reason why it is difficult to accelerate electrons to very high speeds using electric fields. The other reason is the relativistic effects close to light speed.