11.10 Electric Potential | Electricity | Physics 11

Electric potential (often called potential) is a fundamental concept in electrostatics that describes the amount of work energy needed to move a unit of positive charge from a reference point to a specific point in an electric field. It is a scalar quantity that provides a way to characterize the energy of a location within an electric field, independent of the charge being placed there.

Electric Potential Energy vs. Electric Potential

These two concepts are closely related but distinct.

Electric Potential Energy ( $U_{E}$ ): This is the potential energy a specific charge has due to its position in an electric field. It is the work done to move that charge from a reference point to its current location.
Electric Potential ( $V$ ): This is a property of a location in an electric field. It is the electric potential energy per unit charge.

Relationship:

$V = \frac{U _{E}}{q}$

By definition, it is also the work done ( $W$ ) per unit charge:

$V = \frac{W}{q}$

Units of Electric Potential

The SI unit for electric potential is the Volt (V).
One Volt is defined as one Joule of work per Coulomb of charge:

$1 V = 1 J/C$

Calculation of Electric Potential

a) For a Single Point Charge

The electric potential ( $V$ ) at a distance ( $r$ ) from a single point source charge ( $Q$ ) is given by:

$V = k \frac{Q}{r}$

$k$ is Coulomb's constant ( $k \approx 9 \times 1 0^{9} N \cdot m^{2} / C^{2}$ ).
$Q$ is the source charge.
$r$ is the distance from the source charge.
Note: Unlike the electric field (a vector), potential is a scalar. The sign of the charge $Q$ is included in the calculation. A positive charge creates a positive potential, and a negative charge creates a negative potential.

Derived Units→

b) For Multiple Charges

The total electric potential at a point due to a collection of charges is simply the algebraic sum of the potentials created by each individual charge.

$V_{total} = k \sum_{i} \frac{q _{i}}{r _{i}} = k (\frac{q _{1}}{r _{1}} + \frac{q _{2}}{r _{2}} + \dots)$

Potential Difference ( $Δ V$ )

In practice, we are often more interested in the potential difference (or voltage) between two points, as this is what drives the flow of charge (current).

$Δ V = V_{B} - V_{A} = \frac{W _{A B}}{q}$

This is the work required per unit charge to move a charge from point A to point B.

Work done moving a charge in an electric field.

Relationship Between Electric Field and Potential Gradient

The electric field intensity ( $E$ ) at any point is related to the rate of change of electric potential with distance (the potential gradient):

$E = - \frac{Δ V}{Δ r}$

The negative sign indicates that the electric field points in the direction of decreasing potential (from high to low potential). This also gives an alternative unit for electric field: V/m, which is equivalent to N/C.

For uniform fields (e.g., between parallel plates separated by distance $d$ ):

$E = \frac{Δ V}{d}$

Equipotential Surfaces

An equipotential surface is a surface on which the electric potential is the same at every point. No work is done in moving a charge between two points on an equipotential surface, since $W = q Δ V = q \times 0 = 0$ .

Key properties:

Electric field lines are always perpendicular to equipotential surfaces.
For a point charge, equipotential surfaces are concentric spheres.
For a uniform field (parallel plates), equipotential surfaces are parallel planes.

The Electron-Volt (eV)

In atomic and nuclear physics, a more convenient unit of energy is the electron-volt (eV).

Definition: One electron-volt is the amount of energy gained or lost by a single electron when it moves across a potential difference of one volt.
Conversion: $1 eV = (1.602 \times 1 0^{- 19} C) \times (1 V) = 1.602 \times 1 0^{- 19} J$ .
For a particle with charge $q = n e$ accelerated through potential difference $Δ V$ : $KE = n \times Δ V eV$ .

Worked Examples

Example 1: What is the electric potential at a distance of 3 m from a point charge of $5 μ C$ ?

$V = k \frac{Q}{r} = (9 \times 1 0^{9}) \frac{5 \times 1 0 ^{- 6}}{3} = 15 \times 1 0^{3} V = 15 kV$

Example 2: What is the potential difference if it takes 600 J of work to move a 2 C charge between two points?

$Δ V = \frac{W}{q} = \frac{600 J}{2 C} = 300 V$

Example 3: Two parallel plates are separated by 2 cm and have a potential difference of 2 V. Find the electric field between them.

$E = \frac{Δ V}{d} = \frac{2 V}{0.02 m} = 100 V/m$

Electric Potential Energy vs. Electric Potential

These two concepts are closely related but distinct.

Electric Potential Energy ( $U_{E}$ ): This is the potential energy a specific charge has due to its position in an electric field. It is the work done to move that charge from a reference point to its current location.
Electric Potential ( $V$ ): This is a property of a location in an electric field. It is the electric potential energy per unit charge.

Relationship:

$V = \frac{U _{E}}{q}$

By definition, it is also the work done ( $W$ ) per unit charge:

$V = \frac{W}{q}$

Units of Electric Potential

The SI unit for electric potential is the Volt (V).
One Volt is defined as one Joule of work per Coulomb of charge:

$1 V = 1 J/C$

Calculation of Electric Potential

a) For a Single Point Charge

The electric potential ( $V$ ) at a distance ( $r$ ) from a single point source charge ( $Q$ ) is given by:

$V = k \frac{Q}{r}$

$k$ is Coulomb's constant ( $k \approx 9 \times 1 0^{9} N \cdot m^{2} / C^{2}$ ).
$Q$ is the source charge.
$r$ is the distance from the source charge.
Note: Unlike the electric field (a vector), potential is a scalar. The sign of the charge $Q$ is included in the calculation. A positive charge creates a positive potential, and a negative charge creates a negative potential.

Derived Units→

b) For Multiple Charges

The total electric potential at a point due to a collection of charges is simply the algebraic sum of the potentials created by each individual charge.

$V_{total} = k \sum_{i} \frac{q _{i}}{r _{i}} = k (\frac{q _{1}}{r _{1}} + \frac{q _{2}}{r _{2}} + \dots)$

Potential Difference ( $Δ V$ )

In practice, we are often more interested in the potential difference (or voltage) between two points, as this is what drives the flow of charge (current).

$Δ V = V_{B} - V_{A} = \frac{W _{A B}}{q}$

This is the work required per unit charge to move a charge from point A to point B.

Relationship Between Electric Field and Potential Gradient

The electric field intensity ( $E$ ) at any point is related to the rate of change of electric potential with distance (the potential gradient):

$E = - \frac{Δ V}{Δ r}$

For uniform fields (e.g., between parallel plates separated by distance $d$ ):

$E = \frac{Δ V}{d}$

Equipotential Surfaces

Key properties:

Electric field lines are always perpendicular to equipotential surfaces.
For a point charge, equipotential surfaces are concentric spheres.
For a uniform field (parallel plates), equipotential surfaces are parallel planes.

The Electron-Volt (eV)

In atomic and nuclear physics, a more convenient unit of energy is the electron-volt (eV).

Definition: One electron-volt is the amount of energy gained or lost by a single electron when it moves across a potential difference of one volt.
Conversion: $1 eV = (1.602 \times 1 0^{- 19} C) \times (1 V) = 1.602 \times 1 0^{- 19} J$ .
For a particle with charge $q = n e$ accelerated through potential difference $Δ V$ : $KE = n \times Δ V eV$ .

Worked Examples

Example 1: What is the electric potential at a distance of 3 m from a point charge of $5 μ C$ ?

$V = k \frac{Q}{r} = (9 \times 1 0^{9}) \frac{5 \times 1 0 ^{- 6}}{3} = 15 \times 1 0^{3} V = 15 kV$

Example 2: What is the potential difference if it takes 600 J of work to move a 2 C charge between two points?

$Δ V = \frac{W}{q} = \frac{600 J}{2 C} = 300 V$

Example 3: Two parallel plates are separated by 2 cm and have a potential difference of 2 V. Find the electric field between them.

$E = \frac{Δ V}{d} = \frac{2 V}{0.02 m} = 100 V/m$