Electrostatic Potential
Electrostatic Potential in a Uniform Field
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 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, 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.
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.
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.
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 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.
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
Ew = V x q
Where :-
Where :-
Ew - Work Done (Joules)
Ew - Work Done (Joules)
V - Electrostatic Potential (Volts)
V - Electrostatic Potential (Volts)
q - Charge (Coulombs)
q - Charge (Coulombs)
Definition of a Volt
Definition of a Volt
The above formula allows the Volt to defined as :-
The above formula allows the Volt to defined as :-
one Volt (1 V) = one Joule per Coulomb (1 JC -1)
one Volt (1 V) = one Joule per Coulomb (1 JC -1)
The Electron-Volt
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.
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 :-
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 = q x V
E = 1.6 x10-19 x 1
E = 1.6 x10-19 x 1
1 eV = 1.6 x10-19 J
1 eV = 1.6 x10-19 J
Potential Difference
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.
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.
To move a unit positive charge from B to A, work must be done.
Work Done = Force x Distance
Work Done = Force x Distance
Ew = ( E x q ) x d
Ew = ( E x q ) x d
V x q = ( E x q ) x d
V x q = ( E x q ) x d
V = E x d
V = E x d
Where :-
Where :-
V - Potential Difference ( Volts )
V - Potential Difference ( Volts )
E - Electric Field Strength ( Vm-1 )
E - Electric Field Strength ( Vm-1 )
d - distance between points ( m )
d - distance between points ( m )
Note - The above formula can only apply to a uniform field, where E is constant.
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).
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 :-
For non-uniform fields, E can be found by :-
E = - dV/dx
E = - dV/dx
Example 1 -
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?
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
V = E x d
E = V / d
E = V / d
E = 12 / 200x10-6
E = 12 / 200x10-6
E = 6.0x104 NC -1
E = 6.0x104 NC -1
Non Examinable Derivation - Electrostatic Potential due to point charges
Non Examinable Derivation - Electrostatic Potential due to point charges
As stated previously, as the Electric field strength around a point charge varies, so does the Electrostatic Potential.
As stated previously, as the Electric field strength around a point charge varies, so does the Electrostatic Potential.
E = - dV/dr
E = - dV/dr
Also previously, the Electric field strength could also be found using :-
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 :-
By combining the above formulae an expression for Electrostatic Potential ( V ) can be derived :-
Integrating the above gives :-
Integrating the above gives :-
Example 2 -
Example 2 -
What is the net Electrostatic Potential at the point x in the diagram below :-
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 :-
Using the Formula above, applied to each Charge separately and then combined, the net Electrostatic Potential can be found :-
For charge A :-
For charge A :-
Where:-
Where:-
Q = 2x10-6 C
Q = 2x10-6 C
r = 0.25 m
r = 0.25 m
VA = (9x109 x 2x10-6) / 0.25
VA = (9x109 x 2x10-6) / 0.25
VA = 72000 Volts
VA = 72000 Volts
For charge B :-
For charge B :-
Where:-
Where:-
Q = -5x10-6 C
Q = -5x10-6 C
r = 0.75 m
r = 0.75 m
VB = (9x109 x -5x10-6) / 0.75
VB = (9x109 x -5x10-6) / 0.75
VB = -60000 Volts
VB = -60000 Volts
Net Electric Potential at point X = VA + VB
Net Electric Potential at point X = VA + VB
= 12 kV
= 12 kV
Electric Potential Energy
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.
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 :-
The Electric Potential Energy of a charged particle can be found using the following relationship :-
Ep = VQ
Ep = VQ
Where :-
Where :-
Ep = Electric Potential Energy ( Joules )
Ep = Electric Potential Energy ( Joules )
V = Electric Potential ( Volts )
V = Electric Potential ( Volts )
Q = Charge on the particle ( Coulombs )
Q = Charge on the particle ( Coulombs )
Example 3 -
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.
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
Q1 = 5.5x10-9 C
QA = 3.2x10-19 C
QA = 3.2x10-19 C
r = 2.5x10-6 m
r = 2.5x10-6 m
Ep = VQ
Ep = VQ
= (( 9x109 x 5.5x10-9 ) / 2.5x10-6 ) x 3.2x10-19
= (( 9x109 x 5.5x10-9 ) / 2.5x10-6 ) x 3.2x10-19
= 6.3x10-12 J
= 6.3x10-12 J