Fields Tutorial 5 - Electric Potential and Energy

 

Electric Potential, V

When we looked at gravity fields, we saw that gravitational potential energy is the energy needed to bring an object from infinity to a certain point.  We also saw that gravitational potential was the energy per unit mass.  There are two similar quantities in electric fields, potential energy and electric potential.

 

Suppose we moved a charge form one point to another in an electric field.  We would have to do a job of work against the field (or get work out if it were with the field), so there is an energy change.  We can show this as a graph:

 

 

The area under the graph represents the energy transformed.  Using the mathematical trick of calculus, we can show that the energy is given by the equation:

 

 

We will look at the calculus treatment at the end of tutorial.

 

Question 1

A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm.  What is the work done?  What does the sign mean? 

Answer

 

Electric Potential

We can define the electrical potential as the energy per unit charge.  It has the code Ve and is given by the equation:

 

 

The units are Joules per Coulomb (JC-1) or Volts (V).  Indeed the volt is defined as the energy transformed when unit charge is moved between two points.  We can show the relationship between the electric field and potential in this graph.  Potential is the area under the graph:

 

Worked Example

If a hydrogen atom consists of a proton and an electron at an average separation of 5.0 10-11 m, what is the electric potential at this distance and the potential energy of the electron?  What would the field strength be?  Give your answer to an appropriate number of significant figures. (Electronic charge = 1.60 10-19 C)

Answer

Use Ve = __1__ Q = 9 109 m F-1 1.6 10-19 C = 28.8 V = 29 V (2 s.f)

             4pe0   r                5.0 10-11 m

 

Potential energy = charge potential = 28.8 V 1.60 10-19 C = 4.61 10-18 J = 4.6 10-18 J (2 s.f)

Field Strength = potential separation = 28.8 V 5.0 10-11 m = 5.76 1011 N C-1 =  5.8 1011 N C-1 (2 s.f)

 

Equipotentials

All points within an electric field that have the same potential are called equipotentials, rather like contours on a map.  In a radial field, the equipotentials are concentric circles.  In a uniform field, they are parallel lines equally spaced.

 

Suppose we have the positive plate in a uniform field at +300 V, and the negative plate is at 0 V.  Suppose we have 10 equipotentials equally spaced in the uniform field.  Each equipotential has a potential difference of + 30 V.  So there are equipotentials at 0 V, 30 V, 60 V, 90 V, etc.

 

When you travel along a line of equipotential, the work done is zero.

 

The pattern for a radial field around a positive charge is like this:

 

 

Notice how the equipotential lines are at 90 degrees to the field lines.  Notice also that the closer you get towards the charge, the higher the potential (the work done per coulomb) becomes.  Viewed from the side, our potential hill would be something like this:

 

 

 

Question 2

Why are equipotentials not equally spaced in a radial field?

Answer

 

The pattern for the equipotentials for the field between two unlike charges is like this:

 

Image from TAP (IoP)

Note how the equipotential line half way between the two charges is straight.

 

For two like charges, the pattern is like this:

 

Image from TAP (IoP)

Note the pattern of equipotentials around the neutral point.

 

Question 3

A tiny negatively charged oil drop is held stationary in an electric field between two horizontal parallel plated as shown. 

The mass is 2.5 10-10 kg.

 

 

(a)    What are the two forces acting on the drop and in which direction do they act?

(b)   The drop is stationary.  What can be said about the two forces?

(c)    What is the charge on the oil drop?

Answer

 

The pattern for a uniform field is like this:

 

When we move along a line of equipotential, we do no work at all.  Contour lines on a map are lines of equipotential.

 

The equation for the electric field is:


 

When we move between any two equipotentials, we move through a potential difference, or a difference in energy per unit charge.  We have met this before in current electricity.

 

 

Comparing Electric and Gravitational Fields

There are many analogies that can be drawn between electric fields and gravity fields.  Theoretical physicists would go as far as saying that the two are possibly different manifestations of the same thing.  Let us compare the two:

 

Feature

Electric Field

Gravity Field

Quantity susceptible to the force

Charge

Mass

Constant of Proportionality

__1__

 4pe0     

where e0 is the permittivity of free space.  The value of e can be changed by adding a material.

G

 

The value of G, the universal gravity constant has the same value for all media, including a vacuum. 

Relationship with distance

Inversely proportional to r2.

Inversely proportional to r2.

Force Equation

F =  __1__ Q1Q2

         4pe0   r2

F = -G m1m2

              r2

Direction of force

Can be attractive or repulsive

Always attractive

Relative strength

Strong at close range

Weak.  Can only be felt with massive objects

Range

Infinite

Infinite

 

The gravitational attraction between particles in an atom is so small as to be negligible.  The nucleus and its electrons are held together entirely by electrostatic forces, and these are involved in chemical reactions.

 

Gravity forces hold planets together and hold them in their orbits.  Electrostatic forces over the interplanetary distances can be ignored.

 

 

Using Calculus to link Electrostatic Force and Electrostatic Energy (Extension)

The graph below was produced to model the force acting on a positive test charge of 5.0 10-9 C at certain distances from a second charge of 5.0 10-9 C.  The distance varies between 1.0 mm and 1.5 mm.  The graph shows the force against the distance:

We know that if we were to move the test charge from 4.0 10-3 m to 1.0 10-3 m, we would have to do a job of work, as we are applying a force against the repulsive electric field.  On the graph below, the red lines show this and the work done is the area under the graph between the two red lines:

 

The counting of squares would only give an approximation.  Calculus integration would give us a more precise result.  So we can write:

 

We know that:

 

Where:

So we can write this as:

 

In this case k, Q1, and Q2 are constants.  Using the power rule for integration, we can write:

 

Maths Note

Power rule for integration

 

For differentiation the converse is true:

 

Yes, I have missed out the constant, C.  In this case, C = 0.

 

We'll do a worked example.

 

Worked example

A test charge of 5 nC is moved towards a second charge of 5 nC from a position of 4.0 mm to 1.0 mm  Calculate the work that needs to be done. 

Answer

Use:

W = [(8.99 109 m F-1 5.0 10-9 C 5.0 10-9 C) 1.0 10-3 m] - [(8.99 109 m F-1 5.0 10-9 C 5.0 10-9 C) 4.0 10-3 m]

 

W = 2.25 10-4 J - 5.62 10-5 J = 1.69 10-4 J = 1.7 10-4 J (2 s.f.)

 

The positive sign means that work has to be put in. 

The work done results in an increase in potential energy of +1.7 10-4 J.

 

 

We can apply the same rule to the relationship between electrical field strength and electrical potential.  Electrical field strength is force per unit charge and electrical potential is energy per unit charge. This is shown on the graph:

 

We can therefore write:

 

The converse is also true.  The electrical field strength is the gradient of the graph of electrical potential against distance.  So we can write:

 

 

Similarly the force is product of the electrical field strength and the charge Q2, so we can write: