19.2 Electric Potential Due to a Point Charge | Electric Potential And Capacitor | Physics 12

Electric Potential at a Point

The absolute electric potential $V$ at a distance $r$ from an isolated point charge $q$ is defined as the work done per unit positive charge in bringing a test charge from infinity to that point:

$V = \frac{1}{4 π ϵ _{0}} \frac{q}{r}$

where $ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ is the permittivity of free space, and $k = \frac{1}{4 π ϵ _{0}} \approx 9 \times 1 0^{9} N m^{2} C^{- 2}$ .

Key Properties

Scalar quantity: Electric potential has magnitude but no direction. The total potential due to multiple charges is found by simple algebraic (not vector) addition.
Sign: Positive charges produce positive potential; negative charges produce negative potential.
Reference point: The potential is defined to be zero at infinity ( $r \to \infty$ ).
Variation with distance: $V \propto \frac{1}{r}$ (inverse relationship), whereas electric field $E \propto \frac{1}{r ^{2}}$ .

A negative potential (produced by a negative point charge) means that work must be done by an external agent to move a positive test charge from that point to infinity (against the attractive force).

Electric Field as Negative Potential Gradient

The electric field $E$ at a point is related to the rate of change of electric potential with distance:

$E = - \frac{Δ V}{Δ r}$

The negative sign indicates that the electric field points in the direction of decreasing potential — from high potential to low potential. The quantity $\frac{Δ V}{Δ r}$ is called the potential gradient.

Units: The potential gradient has units of $V m^{- 1}$ , which is equivalent to $N C^{- 1}$ .

Worked Example

A point charge $q = + 2 μC$ is located at the origin. Find the electric potential and electric field at $r = 0.3 m$ .

$V = \frac{k q}{r} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{0.3} = 60, 000 V = 60 kV$

$E = \frac{k q}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{( 0.3 ) ^{2}} = 200, 000 V m^{- 1} = 200 kV m^{- 1}$

Note: $E = V / r$ only holds for a point charge (since $V = k q / r$ and $E = k q / r^{2}$ , giving $E = V / r$ ).

Electric Potential Energy of Two Point Charges

The electric potential energy $U$ of a system of two point charges $q_{1}$ and $q_{2}$ separated by a distance $r$ is:

$U = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r} = k \frac{q _{1} q _{2}}{r}$

This represents the work done in assembling the two charges from infinity to a separation $r$ .

If $q_{1}$ and $q_{2}$ have the same sign: $U > 0$ (repulsive system — energy must be supplied to bring them together).
If $q_{1}$ and $q_{2}$ have opposite signs: $U < 0$ (attractive system — energy is released as they come together).

Relationship to Potential

The potential energy can also be written as:

$U = q_{2} V_{1}$

where $V_{1} = \frac{k q _{1}}{r}$ is the potential due to charge $q_{1}$ at the location of $q_{2}$ .

Superposition of Potentials

For a system of $n$ point charges, the total electric potential at a point $P$ is the algebraic sum:

$V_{total} = \sum_{i = 1}^{n} \frac{k q _{i}}{r _{i}}$

Because potential is a scalar, this sum is straightforward — unlike the vector addition required for electric fields.

Example: Square Arrangement

Four identical positive charges at corners of a square: At the centre, $E = 0$ (fields cancel by symmetry) but $V = 4 k q / r \neq = 0$ (potentials add).
Alternating $+ Q, - Q, + Q, - Q$ at corners: At the centre, $E = 0$ (by symmetry) and $V = 0$ (positive and negative contributions cancel).

Electric Potential at a Point

The absolute electric potential $V$ at a distance $r$ from an isolated point charge $q$ is defined as the work done per unit positive charge in bringing a test charge from infinity to that point:

$V = \frac{1}{4 π ϵ _{0}} \frac{q}{r}$

where $ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ is the permittivity of free space, and $k = \frac{1}{4 π ϵ _{0}} \approx 9 \times 1 0^{9} N m^{2} C^{- 2}$ .

Key Properties

Scalar quantity: Electric potential has magnitude but no direction. The total potential due to multiple charges is found by simple algebraic (not vector) addition.
Sign: Positive charges produce positive potential; negative charges produce negative potential.
Reference point: The potential is defined to be zero at infinity ( $r \to \infty$ ).
Variation with distance: $V \propto \frac{1}{r}$ (inverse relationship), whereas electric field $E \propto \frac{1}{r ^{2}}$ .

Negative Electric Potential

Electric Field as Negative Potential Gradient

The electric field $E$ at a point is related to the rate of change of electric potential with distance:

$E = - \frac{Δ V}{Δ r}$

Units: The potential gradient has units of $V m^{- 1}$ , which is equivalent to $N C^{- 1}$ .

Worked Example

A point charge $q = + 2 μC$ is located at the origin. Find the electric potential and electric field at $r = 0.3 m$ .

$V = \frac{k q}{r} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{0.3} = 60, 000 V = 60 kV$

$E = \frac{k q}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{( 0.3 ) ^{2}} = 200, 000 V m^{- 1} = 200 kV m^{- 1}$

Note: $E = V / r$ only holds for a point charge (since $V = k q / r$ and $E = k q / r^{2}$ , giving $E = V / r$ ).

Electric Potential Energy of Two Point Charges

The electric potential energy $U$ of a system of two point charges $q_{1}$ and $q_{2}$ separated by a distance $r$ is:

$U = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r} = k \frac{q _{1} q _{2}}{r}$

This represents the work done in assembling the two charges from infinity to a separation $r$ .

If $q_{1}$ and $q_{2}$ have the same sign: $U > 0$ (repulsive system — energy must be supplied to bring them together).
If $q_{1}$ and $q_{2}$ have opposite signs: $U < 0$ (attractive system — energy is released as they come together).

Relationship to Potential

The potential energy can also be written as:

$U = q_{2} V_{1}$

where $V_{1} = \frac{k q _{1}}{r}$ is the potential due to charge $q_{1}$ at the location of $q_{2}$ .

Superposition of Potentials

For a system of $n$ point charges, the total electric potential at a point $P$ is the algebraic sum:

$V_{total} = \sum_{i = 1}^{n} \frac{k q _{i}}{r _{i}}$

Because potential is a scalar, this sum is straightforward — unlike the vector addition required for electric fields.

Example: Square Arrangement

Four identical positive charges at corners of a square: At the centre, $E = 0$ (fields cancel by symmetry) but $V = 4 k q / r \neq = 0$ (potentials add).
Alternating $+ Q, - Q, + Q, - Q$ at corners: At the centre, $E = 0$ (by symmetry) and $V = 0$ (positive and negative contributions cancel).