Fields Tutorial 4 - Electric Fields



Coulomb’s Law

There are many parallels between gravity fields and electric fields.  The electric field is the region of force around a point charge.  In 1784 Charles Augustine de Coulomb (pronounced "Collume") showed that the force between two point charges was proportional to each charge and inversely proportional to the square of the distance between them.  This reminds us of Newton’s Law of Gravitation. 


Coulomb’s Law can be summarised in code as:

We can rewrite this proportionality as:


The constant k has a value of 8.99 ´ 109 Nm2 C-2 for two charges in a vacuum.  The constant is not normally written as k, but instead:


 [e0 = 8.85 ´ 10-12 C2 N-1 m-2 or F/m]


This term e0 (‘epsilon nought’ - e is a Greek letter ‘e’) is called the permittivity of free space.  We can write the units for k as m/F (metres per Farad).  The equation is written as:


A useful dodge is:



We can approximate the value of k to 9.0 ´ 109 m/F.  Its large value tells us why electrical forces are very strong compared to gravity forces, even when the charges are quite small.  This helps us to explain why solid materials can exist at all.

The force F is positive when the charges are both positive or both negative; this indicates that the force is repulsive, which is borne out by our elementary observations of electrostatics.  If one charge is positive and the other is negative, we can easily see that the force F is negative, therefore attractive.


Question 1

Describe the similarities and differences between the forces in a gravity field and electric field.

Question 2

Find the electrostatic force between a proton and an electron in a hydrogen atom if their separation is 5.3 ´ 10-11 m.  What does the sign tell you?




Electric Forces as Vectors

While gravity fields and forces are attractive, electric forces and fields can be repulsive or attractive.  Force and field are both vector quantities in both situations.  Therefore, if we place a point charge between two other charges we can consider three different situations.

1. Unlike charges in line

In this first situation, the test charge q experiences a force, F1 which is attractive because the charge Q1 is opposite.  At the same time, it also experiences a repulsive force, F2 because Q2 has the same sign.  Therefore:

Fres = F1 + F2

2. Like charges in line

Now, let’s make all the charges positive:

This time the forces are in opposite directions.  Therefore:

F res = F1 – F2

There will be a point at which the forces are equal and opposite, so the test charge stays where it is.


3. Charges not in line

The third situation is when the test charge is NOT in line with the other two charges:

In this case, the resultant force is the vector sum of the two forces, i.e.

(F res)2 = F12 + F22

We can apply the rules for any number of charges around a test charge.

Worked example

A 15 nC charge is placed 150 mm from a test charge of +2 nC.  The third charge of 34 nC is placed 120 mm from the test charge, as shown below:

            (a) What is the resultant force on the test charge?

            (b) At what distance will the resultant force be zero?

(a) Work out the forces F1 and F2:


F1 = 9 × 109 × 15 × 10-9 × 2 × 10-9 = 12 N

            (150 × 10-6)2



F2 = 9 × 109 × 34 × 10-9 × 2 × 10-9 = 42.5 N

            (120 × 10-6)2


            F = 42.5 N – 12 N = 30.5 N to the left.


            (b) F1 = F2  




9 × 109 × 15 × 10-9 × 2 × 10-9 = 9 × 109 × 34 × 10-9 × 2 × 10-9          

             d12                                            d22




9 × 109 × 15 × 10-9 × 2 × 10-9 = 9 × 109 × 34 × 10-9 × 2 × 10-9 =       

            d12                                            d22


d22 ÷ d12 = 34 ÷ 15 = 2.27


d2 = √2.27 × d1 = 1.51 × d1


            We also know that d1 + d2 = 150 + 120 = 270 mm



d1 = (270 – d2)




d2 = 1.51 × (270 – d2)


d2 + 1.51 d2 = 406


d2 = 406 ÷ 2.51 = 162 mm



d1 = 270 – 162 = 108 mm



Electric Field Strength

A gravity field is where a force is exerted on a mass; an electric field is a region where a force is exerted on a charge.


The electric field strength is defined as force per unit charge.  In equation form this is represented as:


E = F/Q    

[E – electric field strength; F – force; Q – charge]


E is a vector quantity and its units are newtons per coulomb, N/C.  It can be negative (attractive) or positive (repulsive).


Question 3

 What does it mean when the force per unit charge is                                               










Since the electric field strength is defined as force per unit charge, we can write:

This equation is true for all point charges, which can be considered to have a radial field.  A spherical charge can also be considered to have a radial field provided the distance from the charge to the centre is much greater than the radius of the sphere.



Question 4

From the diagram how can you tell the value of the electric field strength?


Question 5

A charge of +1.6 ´ 10-19 C has a force of 8.7 ´ 10-15 N exerted on it when it is placed a certain point in a radial electric field.  What is the electric field strength?




Electric Field Strength as a Vector


While gravity fields and forces are attractive, electric forces and fields can be repulsive or attractive.  Force and field are both vector quantities in both situations.  Therefore, if we place a point charge between two other charges we can consider three different situations.

1. Unlike charges in line

Since E = F/q, we can say that:


F = F1 + F2 = qE1 + qE2


E = F = qE1 + qE2

  q             q


2. Like charges in line



We can use a similar argument to the one above:

E = F = qE1 - qE2

  q             q


If all the charges were negative, the same would apply.


3. Charges not in line

E2 = E12 + E22




We can plot a graph of electric field strength against distance:



We can see that the graph shows a classic inverse square law.


Question 6

Two point charges  are placed at a distance of 8.0 ´ 10-3 m apart as shown in the diagram.



(a)    Copy the diagram, and show the complete electric field you would expect to see.

(b) What is the force acting between the two points? What can you say about the force?

(c) What is the value of the electric field strength at the midpoint between the charges?



The direction of the field is the direction of the force.  The force in this case is the force exerted by the electric field on a hydrogen ion (proton).


If we are looking at the electric field between two point charges, we need to take both into account both the charges.




Uniform Electric Fields


Where two charged plates are close together, the radial fields of the charges combine to make a uniform field.



Notice that the field bulges at the ends; generally we ignore this.


In this case we can show that the electric field strength is given by a simpler relationship:


E = V/d          

[E – electric field strength; V – voltage; d distance in m]


Strange as it may appear, the units of volts per metre are entirely consistent with newtons per coulomb.


Question 7

Two plates are placed 15 cm apart connected to a 4500 V supply.


(a)    What is the strength of the uniform field between the plates?

(b)     What is the force acting on a 6 nC test charge between the plates?




Trajectory of charged particles in a uniform electric field


Charged particle such as electrons behave in exactly the same way as projectiles do in a gravity field.  You may want to revise the ideas from AS Physics 2.  You will remember that:

  • The vertical and horizontal movements are independent.

  • The horizontal velocity stays the same.

  • The vertical velocity changes because of the acceleration due to the gravity field.

Exactly the same applies with electric fields.



The electron is attracted by the positive plate:

  • Its horizontal velocity is not changed at all.

  • Its vertical velocity starts at zero as it enters the field, and increases to vy (in an upwards direction) as it leaves the field.

  • The path is a parabola.

Unlike a stone that is thrown horizontally that will hit the ground, once an electron leaves the electric field, it will continue in a straight line.  It will have a resultant velocity that is the vector sum of the vertical and horizontal velocities.


Use the equations of motion to analyse the two different movements.

Let us look at an example:


The electric field strength between a pair of plates length 4.0 cm in a cathode ray tube has a value 23 000 N/C.  An electron enters the field at right angles at a velocity of 3.7 × 107 m/s from left to right.  What is the velocity and direction of the electron as it leaves the field?


Before we attempt our answer, we need to think about how the electron is going to travel.  It is travelling in a straight line as it enters the electric field.  As it goes in, there will be a force acting on the electron that attracts it to the negative plate.  This will cause the electron to accelerate towards the positive plate.  Since its horizontal velocity is unchanged and its vertical velocity is increasing, it describes a parabolic path.  (If the plates were long enough, or if the electron were slow enough, the electron would eventually hit the positive plate.)  The electron leaves the field and travels again in a straight line, but this time at an angle to its un-deviated path.  We can work out the resultant velocity by consideration of the horizontal and vertical velocities.

Horizontal velocity remains unchanged at 3.7 ´ 107 m/s, since there is no horizontal force.

We can now work out the upwards force using F = QE


F = -1.6 ´ 10-19 ´ 23000 = -3.69 ´ 10-15 N


The minus sign tells us that the force is attractive, and is against the direction of the field, which is from positive to negative.  However we will ignore the minus sign from now on.


We now need to work out the upwards velocity of the electron, which we do by multiplying the acceleration by the time.  We need to know the acceleration and the time.  Simple enough, really.


Time taken to travel through the field = distance ¸ horizontal speed

                                                             = 0.04 m ¸ 3.7 ´ 107 m/s

                                                             = 1.08 ´ 10-9 s


Then we use Newton II to find acceleration, a = F/m.


a = 3.69 ´ 10-15 N ¸ 9.1 ´ 10-31 kg = 4.04 ´ 1015 ms-2.

Upwards velocity = at = 4.04 ´ 1015 ms-2 ´ 1.08 ´ 10-9 s = 4.37 ´ 106 m/s


Now we can do the vector addition to work out the resultant velocity

v2 = (3.7 ´ 107 m/s)2 + (4.37 ´ 106 m/s)2 = 1.388 ´ 1015 m2s-2

Þ v = 3.73 ´ 107 m/s

Now we can work out the angle of deflection, q.


q = tan-1 4.37 = tan-1 0.118 = 6.7 o



The velocity is 3.73 ´ 107 m/s in a direction of 6.7 o to the horizontal.





The charged particle can, of course, be positive.  A proton can pass through an electric field, as can an alpha particle, or a positive metal ion.


Question 8

An alpha particle enters a uniform electric field as shown:

The plates are 5 cm long.

(a) Calculate the electric field strength.

(b) Calculate the mass of the alpha particle (mass of a nucleon = 1.67 × 10-27 kg)

(c) What is the charge on an alpha particle?

(d) Calculate the force on the alpha particle.

(e) Calculate the acceleration of the alpha particle.

(f) Calculate the time taken for the alpha particle to pass between the plates.

(g) Calculate the vertical velocity.

(h) Explain whether there is a noticeable deflection.  Justify your answer.