Electrostatic Potential

Electrostatic Potential in a Uniform Field

In an Electric field diagram, the Electric field strength is shown by the separation of the field lines. The closer together the field lines are, the stronger the Electric field is at that point :-

In the above diagram, the field lines are closest together near to the point charges, giving a strong field at these points.

In the above diagram, as the field lines all have equal spacing, the Electric field strength is uniform between the plates.

Note - the field is only uniform within the plates. At the edges, the field lines curve, giving a non-uniform field. This is called the edge effect. 

Earlier, it was shown that Electrostatic Force acted in a similar way to Gravitational Force. The same can be shown between Gravitational Potential and Electrostatic Potential. 

The Gravitational Potential describes how much work must be done per unit mass to move an object in that field. The Gravitational potential at a point is defined as the work done to bring a unit mass from infinity to that point. 

The Electrostatic Potential (V) describes how much work must be done per unit positive charge to move that charge in an Electric field. The Electrostatic Potential at a point is defined as the work done to bring a unit positive charge from infinity to that point. 

Ew  =  V x q 

Where :-

Ew - Work Done (Joules) 

V - Electrostatic Potential (Volts) 

q - Charge (Coulombs)


Definition of a Volt

The above formula allows the Volt to defined as :-

one Volt (1 V)  =  one Joule per Coulomb (1 JC -1)


The Electron-Volt

In Particle Accelerators, the Energy associated with collisions have values that are not given in Joules, but in Electron-Volts (eV).  An Electron-Volt is a unit of Energy measurement that is used in high Energy particle Physics, as working in Joules results in values that a so small that they become meaningless. 

1 Electron-Volt (1 eV) is defined as the Work Done to move a single Electron through a Potential Difference of 1 Volt. By calculation, the Energy required is found by :-

E  =  q x V 

E  =  1.6 x10-19  x  1

1 eV  =  1.6 x10-19

Potential Difference

The Potential Difference (V) between two points A and B (separated by distance d)  is defined as the Work Done to move a  unit positive charge between those points. 

To move a unit positive charge from B to A, work must be done. 

Work Done  =  Force x Distance

Ew  =  ( E x q ) x d 

V x q  =  ( E x q ) x d 

V  =  E x d 

Where :-

V - Potential Difference ( Volts )

E - Electric Field Strength ( Vm-1

d - distance between points ( m ) 

Note - The above formula can only apply to a uniform field, where E is constant. 

Also, the units of Electric field strength are different from what has been shown before, implying that NC -1 and Vm-1 are equivalent to each other. Therefore the Electric field is a  measure of the Potential gradient (How V changes with distance).  

For non-uniform fields, E can be found by :-

E  = - dV/dx

Example 1 - 

The Potential Difference between two plates of a charged parallel plate capacitor is 12 V. What is the Electric field strength between the plates if their separation is 200x10-6 m?

V  =  E x d 

E  =  V / d

E = 12 / 200x10-6 

E  =  6.0x104 NC -1 

Electrostatic Potential due to point charges 

As stated previously, as the Electric field strength around a point charge varies, so does the Electrostatic Potential. 

E  = - dV/dr

Also previously, the Electric field strength could also be found using :-

By combining the above formulae an expression for Electrostatic Potential ( V ) can be derived :-

Integrating the above gives :-

Example 2 - 

What is the net Electrostatic Potential at the point x in the diagram below :-

Using the Formula above, applied to each Charge separately and then combined, the net Electrostatic Potential can be found :-

For charge A :-

Where:-

                 Q  =  2x10-6 C

                  r  =   0.25 m 

                

VA  =  (9x109 x 2x10-6) / 0.25 

                VA =  72000 Volts

For charge B :-

Where:-

                 Q  =  -5x10-6 C

                  r  =   0.75 m 

          

VB  =  (9x109 x -5x10-6) / 0.75 

      

                VB =  -60000 Volts

Net Electric Potential at point X  =  VA + VB

                                                      =  12 kV

Electric Potential Energy

A charged particle a a given point in an Electric field will have Electric Potential Energy Ep because work must be done moving the charge from infinity to that point. 

The Electric Potential Energy of a charged particle can be found using the following relationship :-

Ep  =  VQ 

Where :-

Ep  = Electric Potential Energy ( Joules )

V  =  Electric Potential ( Volts )

Q  =  Charge on the particle ( Coulombs ) 

Example 3 -

An Alpha particle is 2.5x10-6 m from a point charge of 5.5 nC. Calculate the Potential Energy of the Alpha particle at this position. The charge on the Alpha particle is 3.2x10-19 C.

Q1 = 5.5x10-9 C

QA = 3.2x10-19 C

r =  2.5x10-6 m

Ep   =  VQ

        =  (( 9x109 x 5.5x10-9 ) / 2.5x10-6 ) x 3.2x10-19

        = 6.3x10-12 J