Electric potential

Electric potential

Voltage

Charged particles exert forces on each other.  The electric field
E
= F/q produced by a charged particle at some position
r in
space is a measure of the force F the particle exerts on a test charge q,
if we place the test charge at r.  The electric field E is a
vector.  The electric field due to a charge distribution is the vector
sum of the fields produced by the charges making up the distribution.  When
we think about electricity in everyday life, we seldom think about the electric
fields.  We are more familiar with the concepts of voltage, current, and
power.

How is the voltage related to the electric field?

When a particle with charge q is placed in an external electric field
E,
(i.e. an electric field produced by other charges), then an electric force
F
= qE will act on it.  If this force is not balanced by other forces,
then the particle will accelerate, and its kinetic energy will change.  The
electric force will do positive work on the particle.  If the electric
force is balanced by another external force Fext = -qE,
and if this external force moves the particle against the electric force, than the
external force Fext does positive work. 

Recall!

• The work W done on an object by a constant force is defined as W = F∙d = F d cosθ. 
  Here F is the magnitude of the force, d is the magnitude of the displacement
  vector, and θ is the angle between the directions of the force and displacement vectors.
  The work done by a varying force in one dimension is defined as W = Σxixf
  Fx∆x, as ∆x becomes infinitesimally
  small.  In three dimensions we write
  W = Σ F∙∆r = Σxixf
  Fx∆x +  Σyiyf Fy∆y + Σzizf
  Fz∆z.
   

Therefore the work done done by an external force balancing the electric force is

W = Σ Fext∙∆r = -q Σ
E∙∆r.

In the sums we always assume that the displacements become infinitesimally
small.

The work done by the external force Fext = -qE is equal to the change in the
electrostatic potential energy of the
particle in the external field.  The change in the potential energy of a
charge q when being moved from point A to point B, is the work done by Fext in moving the charge.

∆U = UB – UA = -q ΣAB
E∙∆r.

The sum is taken along a particular path.  But the electrostatic force
is a conservative force.  The work is
independent of the path.  ∆U therefore depends
only on the endpoints A and B of the path, not on the actual path itself.

Embedded Question 1

(a)  Which requires zero work, to move a positive point charge from
point P to point 1, 2, 3, or 4?   All those points are the same
distance from P.
(b)  Which requires the most positive work done by an external force, to
move a positive point charge from point P to point 1, 2, 3, or 4?

Discuss this with your fellow students in the discussion forum!
Review the concept of work in physics.
Discuss the relationship between the work done by and against the electric field
and the electrostatic potential energy.

The change in potential energy is proportional to the charge q.  Its sign
depends on whether the charge is positive or negative.  We define the
potential difference or
voltage ∆V as the
potential energy difference divided by the charge, or the potential energy
difference per unit (positive) charge.  The potential difference is the potential
energy difference of a small, positive test charge, divided by the charge. 
As in the case of gravity, the zero of the potential energy and therefore the
zero of the potential are not uniquely defined.  Only potential energy
differences and potential differences are unique.  The SI unit of energy is
J, therefore the SI unit of potential is J/C.  We define
Volt as V = J/C.

The electrostatic potential is a
scalar, not a vector.

We say that a charge distribution, which produces an electric field, also
produces an electric potential.  The electrostatic potential produced by a
finite charge distribution is, by convention, set to zero at infinity. 
Then the potential V(r) of the distribution is the work done per unit
charge in bringing a small test charge from infinity to position
r. 
For a charge distributions which extend to infinity, we cannot set the zero of
the potential at infinity, because then the potential would be infinite
everywhere, and it would be a useless concept.  We then set the zero at
some convenient reference point, but we always must specify the reference point
along with the potential, since there is no unique convention.  For many
applications we choose the ground to be the zero of the potential.

The potential difference between the two terminals of an A, B, C, or D cell
battery is 1.5 V.  For every Coulomb of negative charge that is moved from
the positive to the negative terminal, 1.5 J of work must be done against
electric forces, and 1.5 J of some other form of energy is converted into
electrostatic energy.  When the terminals are connected by a wire, then the
charge is free to move inside the wire, and the electric field does work on the
charge.  This work can now be reconverted into some other form of energy.

The potential at a point r of a positive point charge located at the
origin is the work that must be done per unit charge in bringing a test charge
from infinity to r.

V(r) = -Σ∞r
E∙∆r
= Σr∞ E∙∆r .

We can bring the test charge along an arbitrary path, which we can think of
as being made up of infinitesimal small steps, either in the radial direction or
perpendicular to the radial direction.  We do not have to do work when we step
perpendicular to the radial direction, because then E is perpendicular to
∆r.  Along the radial direction E∙∆r = E∆r, because
E and r points both point outward.  The sum becomes

V(r) = q/(4πε0r) = keq/r.

This expression also gives the potential due to any spherically symmetric
charge distribution outside the distribution.

The potential outside a spherically symmetric charge distribution with total
charge q is the same as that of a point charge q, V(r) = keq/r.

It is is proportional to the inverse of the distance from the point charge.

To find the potential due to a collection of charges, we use the principle of
superposition and add the potentials due to the individual charges.  Because the
potential is a scalar, and not a vector, we just have to add numbers.  To
find the total electrostatic potential energy of a collection of point charges,
sum over all pairs.

Problem:

The
charges in the groups A and B below are all given in units of Q. Which group of
charges took less work by an external force to bring together from infinity?

Solution:

• Reasoning:
  For case A:
  We have only one pair of charges.
  W = ke 2Q*Q/d = 2 keQ2/d.

  For case B:
  We have 3 pairs of charges.
  Bring the two lower charges together: W = ke Q2/d.
  Bring the top charge:
  Do work against the force exerted by the two lower charges
  W = ke Q2/d + ke Q2/d = 2 ke Q2/d.
  Total Work: W = 3 keQ2/d = sum over all pairs.

  Group A took less work by an external force to bring together from infinity
  than group B.

Units:

1 electron volt (eV)  = 1.6*10-19 J.

1 eV is the change in potential energy of a particle with charge qe = 1.6*10-9
C  when the change in potential is 1 Volt (V).  The unit was defined so that
when you know the voltage between two points in space, you know the change in
potential energy of an elementary particle when it moves from one to the other
point.  (The sign of the change in potential energy depends on the sign of the
charge.)  When a free proton moves through a potential difference of 1 V
its kinetic energy decreases by -qV = (1.6*10-19 C)*(1 J/C) = -1.6*10-19 J
= -1 eV. 
When a free electron moves through the same potential difference of 1 V its
kinetic energy increases by -qV = -(-1.6*10-19 C)*(1 J/C) = 1.6*10-19 J
= 1 eV.

Problem:

Assume
that the potential difference between the positive and negative plate is 100 V. 
A proton travels from the positive to the negative plate.  What happens to
the proton’s kinetic energy?

Solution:

• Reasoning:
  The electrostatic force is a conservative force.
  ∆KE + ∆U = 0.
  (change in kinetic energy + change in potential energy = 0.)
• Details of the calculation:
  The proton accelerates towards the negative plate.  Its potential
  energy decreases by 100 eV.
  Its kinetic energy increases by 1.6*10-17 J = 100 eV.

Summary:

By definition, the potential difference is the potential energy difference of
a small, positive test charge, divided by the charge. The zero of the potential
energy and therefore of the potential is not uniquely defined, but chosen at a
convenient place.  Voltage is just another word for potential difference,
or potential energy difference divided by the charge.
∆V = ∆U/q.

A particle always accelerates towards the position with the lower potential
energy U.  The electrostatic potential energy is U = qV, were V is the
potential. Consider a particle with charge of magnitude qe, for example a
proton (+qe) or and electron (-qe).

If q = qe, then U = qeV.  U gets more positive or
higher, the bigger V.  The positively charged particle accelerates towards
the region of lower potential.

If q = -qe, then U = -qeV.  U gets more negative
or lower, the bigger V.  The negatively charged particle accelerates
towards the region of higher potential.

If the potential difference is V, the change in potential energy of the
particle U can be calculated in SI units.  But the nice thing about the
unit eV is, that if the voltage difference is given in volts V, then the change
in the potential energy of the particle, ±qeU, is just the same number
in units of eV.

Equipotential lines

Just as we described the electric field around a charged object by field
lines, we can also describe the electric potential pictorially with
equipotential surfaces (contour plots).  Each
surface corresponds to a different fixed value of the potential.  Equipotential
lines are lines connecting points of the same potential.  They often appear on
field line diagrams. 

Equipotential lines are always perpendicular to field lines.

The electric potential at a position r is the electric potential
energy a charge q has if it is at position r.  As the name
equipotential implies, the potential energy of a charge would be the same
anywhere on an equipotential surface.  If the charge moved along an
equipotential surface, its electric potential energy would not change, no work
would be done by or against the electric field.  But if there was a
component of the electric field tangent to the equipotential surface, then, if
the charge moved along that component, the electric field would do work and the
potential energy of the charge would change.  This is a contradiction. We
cannot have a component of the electric field tangent to an equipotential
surface.  Hence, any electric field must be perpendicular to any
equipotential surface.
 

Example:

Field lines and equipotential lines for a positive point charge are shown
in the figure.  This is a 2-dimensional representation, a cut through the
3-d surfaces.
Please also explore this 3-dimensional representation
below.  
Please click on the image!

Problem:

In the Bohr model of the hydrogen atom, the electron orbits the proton
at a distance of r = 5.29*10-11 m.  The proton has charge +qe
and the electron has charge -qe, where qe = 1.6*10-19
C.  How much work must be done
to completely separate the electron and the proton

Solution:

• Reasoning:
  The force on the electron is the Coulomb force between the proton and
  the electron.  It pulls the electron towards the proton.  For the electron
  to move in a circular orbit, the Coulomb force must equal the
  centripetal force.  We need keqe2/r2
  = mv2/r.
• Details of the calculation:
  mv2
  = keqe2/r, so the
  kinetic energy of the electron is
  KE(r) = ½mv2 = ½keqe2/r.
  
  The potential energy of the electron in the field of the positive proton
  point charge is U(r) = -qeV(r) = – keqe2/r. 
  
  The
  total energy is the sum of the electron’s kinetic energy and its
  potential energy.
  KE(r) + PE(r) = -½keqe2/r = (-½) (9*109)(1.60*10-19)
  /(5.29*10-11) J = -2.18*10-18 J.
  This is usually stated in energy units of electron volts (eV). 
  1 eV =
  1.60*10-19 J.
  -2.18*10-18 J * 1eV/(1.60*10-19
  J) = -13.6 eV.
  To remove the electron from the atom, i.e.
  to move it very far away and give it zero kinetic energy, 13.6 eV of
  work must be done by an external force. 
  13.6 eV is the ionization energy
  of hydrogen.

Problem:

An alpha particle containing two protons is shot directly towards a
platinum nucleus containing 78 protons from a very large distance with a
kinetic energy of 1.7*10-12 J.  What
will be the distance of closest approach?

Solution:

• Reasoning:
  The electric charge of the alpha particle is q1 = 2qe and
  that of the platinum nucleus is q2 = 78 qe.  The
  alpha particle and the nucleus repel each other.  As the alpha particle
  moves towards the nucleus, some of its kinetic energy will be converted into
  electrostatic potential energy.  At the distance of closest approach,
  the alpha particle’s velocity is zero, and all its initial kinetic energy
  has been converted into electrostatic potential energy.
• Details of the calculation:
  The initial kinetic energy of the alpha particle is KE = 1.7*10-12
  J.
  
  The final potential energy is U = kq1q2/d, where d
  is the distance of closest approach.
  We have keq1q2/d = 1.7*10-12
  J.  This yields
  d = keq1q2/(1.7*10-12
  J)
  = (9*109*2*78*(1.6*10-19)2/(1.7*10-12)) m
  = 2.1*10-14 m
  for the distance of closest approach.  The alpha particle’s velocity is
  zero at d.
  The acceleration is in a direction away from the nucleus.  The
  distance between the alpha particle and the nucleus will increase again,
  converting the potential energy back into kinetic energy.

Problem:

An evacuated tube uses a voltage of 5 kV to accelerate electrons from rest to
hit a phosphor screen.
(a)  How much kinetic energy does each electron gain?
(b)  What is the speed of an electron when it hits the copper plate?

Solution:

• Reasoning:
  1 eV is the change in potential energy of a particle with charge qe = 1.6*10-9
  C  when the change in potential is 1 Volt (V).
• Details of the calculation:
  (a)  Each electron loses 5 keV of potential energy and
  gains 5 keV = (5000 eV)(1.6*10-19 J/eV) =
  8*10-16 J of kinetic energy.
  (b)  E = ½mev2.  v2 = 2E/me
  = (2*8*10-16 J)/(9.1*10-31 kg) = 1.75*1015
  (m/s)2.
  v = 4.2*107 m/s.  This is more than 1/10 the speed of light.

The electrostatic potential V is related to the electrostatic field
E.  If the electric field E
is known, the electrostatic potential V can be obtained using V(r) = -Σ∞r
E∙∆r.

How can we obtain the field from the potential?

Consider the two points P1 and P2 shown in figure
above.  Assume that they are separated by an infinitesimal distance ∆L.  The change in the electrostatic potential between P1 and P2
is given by ∆V = –E∙∆L = -E ∆L cosθ.  
Here θ is the angle between the direction of the electric field and the
direction of the displacement vector ∆L.  We can rewrite
this equation as

∆V/∆L = -E cosθ = -EL.
EL indicates the component of the electric field along the
direction of ∆L.  If the direction of the displacement is chosen to
coincide with the x-axis, this becomes

∆V/∆x = -Ex.
For the displacements along the y-axis and z-axis we obtain ∆V/∆y = -Ey
and ∆V/∆z = -Ez.
The total electric field
E can be obtained from the electrostatic potential V
by combining these three equations. 
We say that E is the
negative gradient
of the potential V.

In many electrostatic problems the electric field due to a certain charge
distribution must be evaluated.  The calculation of the electric field can be
carried out using two different methods:

• The electric field can be calculated by applying Coulomb’s law and vector addition of the contributions from all charges in the charge
  distribution.
• The total electrostatic potential V can be obtained from the algebraic
  sum of the potential due to all charges that make up the charge
  distribution, and the field can be found by calculating the gradient of V.

In many cases the second method is simpler, because the calculation of the
electrostatic potential involves an algebraic sum, while the direct calculation
of the field involves a vector sum.

Problem:

The potential difference between the two plates of the capacitor shown below is
12 V.  Equipotential surfaces are shown.  If the separation between the
plates is 1 mm, what is the strength of the electric field between the plates?

Solution:

• Reasoning:
  Let the y-axis point upward.  V only varies with y.  ∆V/∆x = ∆V/∆z
  = 0.  E = Ey j = -∆V/∆y  j.
• Details of the calculation:
  Let the y-axis point upward.  The field is uniform and
  points in the negative y-direction. 
  ∆V/∆y = -Ey
  = (12 V)/(10-3 m).  Ey = -12000 V/m.
  The strength of the electric field between the plates is E = 12000 V/m =
  12000 N/C.

Problem:

The
graph on the right shows a contour map of the equipotential surfaces due to 3
point charges.  Estimate the magnitude and direction of the electric field
at point P.

Solution:

• Reasoning:
  Point P lies between two equipotential surfaces.  The surface above P
  is at 40 V and the surface below is at 20 V.  The electric field points
  away from the positive charge, from higher to lower potential.  It
  points downward at P.
  Given the distance scale, I estimate the perpendicular distance between the
  40 V and 20 V equipontential surfaces near point P to be 2.5 cm.
• Details of the calculation:
  |E| = ∆V/∆y = 20 V/ 2.5*10-2 m = 800 V/m.
  The magnitude on of the electric field at point P is ~800 V/m and it points
  downward.

If you miss having regular lectures, consider this video lecture

Lecture 4: 
Electrostatic Potential, Electric Energy, eV, Conservative Field, Equipotential
Surfaces