OCR A-Level Physics/Fields, Particles and Frontiers of Physics/Electric Fields

Electric fields are produced by any electrically charged object, such as an electron or a proton, or between two oppositely charged plates. The laws behind them form a fundamental part of the A2 physics course.

Electric field strength

The electric field strength of the object is defined as the electrical force experienced per unit charge by a positive charge. It is usually measured in NC^-1, with Vm^-1 being another acceptable equivalent unit. Electric field strength is a vector quantity.

The defining equation for Electric Field Strength (E) is given by:

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

Electric field lines

An electric field can be represented by electric field lines. They always point towards a region of more negative charge: that is, they show the direction in which a positive charge placed at that point would move. The density of the electric field lines indicates the electric field strength, and their arrangement indicates the nature of the electrical field. For example, parallel electric field lines which are evenly spaced show a uniform electric field.

The electric field strength can also be shown by lines of equipotential, which are always at right angles to the electric field lines.

Coulomb's Law

While the electrical force has an infinite range due to the fact that it is carried by photons (a massless particle), the force does decrease significantly with increasing distance. This fact is stated quantitatively by Coulomb's Law, which says that the electrical force exerted by two charged particles on each other is directly proportional to their charges multiplied together and inversely proportional to the square of the distance between two charges.

If we take Q and q to be the charges measured in Coulombs of the charged particles charged particle, and r to be their separation in metres, we can say that $F\propto Qq$ and $F\propto 1/r^{2}$. By combining these two equations together, it can be deduced that $F\propto Qq/r^{2}$, where the constant of proportionality is given as ${\frac {1}{\epsilon _{0}4\pi }}$, with $\epsilon _{0}$ representing the permittivity of free space (8.85*10^-12 Fm^-1).

The overall equation, therefore, is:
$F={\frac {Qq}{\epsilon _{0}4\pi r^{2}}}$

Calculating Electric Field Strength

From the definition of electric field strength, it is known that E=F/Q, so F=EQ. This can substituted into the equation for Coulomb's law to deduce that $EQ={\frac {Qq}{\epsilon _{0}4\pi r^{2}}}$. Dividing both sides of the equation by Q will leave us with the following equation for electric field strength:

$E={\frac {Q}{\epsilon _{0}4\pi r^{2}}}$

The Electric Field Between Parallel Plates

Between oppositely charged parallel plates the electric field will always be uniform, except for at the edges, where the field lines will start to curve. The electric field strength in this uniform field can be calculated by $E={\frac {V}{d}}$, where V is the potential difference between the plates and d is the separation between them.

The derivation for this is as follows:

• Work done = Force * distance $(W=Fd)$, and Voltage = Work done/Charge $\left(V={\frac {W}{Q}}\right)$.
• The second of the above equations can be rearranged into $W=VQ$
• The two equations can be equated, showing that $Fd=VQ$
• Dividing both sides by dQ leaves F/Q = V/d, where F/Q is, by definition, equal to the electric field strength (E)
• Therefore, E = V/d

The motion of particles in a uniform electric field

The direction of the force in an electric field is completely independent of the direction in which the particle is travelling, unlike with magnetic fields. Therefore, the motion of a particle in an electric field will be parabolic, resembling the motion of a mass in a gravitational field.

The component of velocity of the charged particles at right angles to the direction of the electric field will remain constant. However, the component of velocity parallel to the magnetic field will increase, with positive particles being accelerated towards the negative plate and negative ones towards the positive plate.

By taking a simple example of a charged particle travelling perpendicular to an electric field formed by two parallel plates, we can determine the final vertical velocity of the electron the moment it leaves the electric field. To do this, we need to separately form expressions in the horizontal and vertical directions.

Horizontal Component:

$t={\frac {L}{V_{H}}}$, where $V_{H}$ is the horizontal velocity and $L$ is the length of the parallel plates

Vertical Component:

$V_{V}=u+at$, where $V_{V}$ is the vertical velocity

We will take the initial velocity of the charged particle as zero, u = 0

$a={\frac {F}{m}}={\frac {EQ}{m}}$

Combining the two equations above gives:

$V_{V}=\left({\frac {EQ}{m}}\right)\times t$

Finally, we can combine the vertical and horizontal components into a single expression:

$V_{V}={\frac {EQL}{mV_{H}}}$

Electric potential

Electric potential at a point is defined as the work done per unit charge in moving a positive charge from infinity to that point.

If the test charge has a charge $q$, the equation for $V$ can be determined:

$V={\frac {E}{q}}={\frac {Qq}{4\pi \varepsilon _{0}r}}\times {\frac {1}{q}}={\frac {Q}{4\pi \varepsilon _{0}r}}$

The unit for electric potential is $JC^{-1}$ or volts ($V$)

Comparison between electric and gravitational fields

The OCR syllabus also expects an understanding of the similarities and differences between the electric fields discussed in this module and the gravitational fields considered in unit G484.

Similarities

• Both forces involve 'action at a distance'. That is, a mass or charged particle need not be in contact with an object in order to exert an electric or gravitational force on it
• A radial field is produced by both a point mass and point charge
• The field strength is inversely proportional to separation squared and defined by force per unit something

Differences

• Electric forces are produced by charge, while gravitational forces are instead produced by masses
• Gravitational forces can only be attractive (as mass is a scalar), while electric forces can be attractive or repulsive (as charge as a vector)