Electric Potential in Electromagnetism

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Electric Potential in electromagnetism

Electric potential is a fundamental concept in electromagnetism that describes the amount of electric potential energy a unit charge possesses at a specific point in an electric field. It represents the work needed to move a positive test charge from a reference point, typically infinity, to that location without producing acceleration. Measured in volts (V), electric potential helps explain how charges interact, how electric circuits function, and how energy is transferred within electrical systems. Understanding electric potential is essential for studying electric fields, capacitors, and voltage in practical applications ranging from electronics to power distribution.

1.0Definition of Electric Potential

The potential energy per unit charge $U / q_{0}$ is independent of the value of $q_{0}$ and has a value at every point in an electric field. This quantity $U / q_{0}$ is called the electric potential (or simply the potential) V. Thus, the electric potential at any point in an electric field is:

$V = U / q_{0}$

Since potential energy is a scalar quantity so the electric potential also is a scalar quantity. If the test charge is moved between two positions A and B in an electric field, the charge-field system experiences a change in potential energy. Potential is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge $q_{0}$ .

V=Uq0=-ABE.dl

2.0Physical Interpretation of Electric Potential

In an electrostatic field, the electric potential (due to some source charges) at a point P is defined as the work done by an external agent in taking a unit point positive charge from a reference point (generally taken at infinity) to that point P without changing its kinetic energy.

$Δ V = \frac{Δ U}{q _{0}} = - \int_{A}^{B} E \cdot d l$

3.0Mathematical Representation of Electric Potential

If $(W_{\infty \to P})_{(e x t)}$ is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is

$V_{P} = [\frac{( W _{\infty \to P} ) _{e x t}}{q}]_{Δ K = 0} = \frac{( - W _{e l c} ) _{\infty \to P}}{q}$

Note:

(1) $(W_{\infty \to P})_{(e x t)}$ can also be called as the work done by external agents against the electric force on a unit positive charge due to the source charge.

(2) Write both W and q with proper signs.

4.0Properties of Electric Potential

Potential is a scalar quantity, its value may be positive, negative or zero.
SI unit of Potential is $volt = \frac{Joule}{Coulomb}$ and its dimensional formula is $[M^{1} L^{2} T^{- 3} I^{- 1}]$ .
Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point.
Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinity. $(Taking V_{\infty} = 0)$
Potential decreases in the direction of the electric field.
By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

$V_{1} = \frac{1}{4 π ϵ _{0}} \frac{q _{1}}{r _{1 P}}, V_{2} = \frac{1}{4 π ϵ _{0}} \frac{q _{2}}{r _{2 P}}, V_{3} = \frac{1}{4 π ϵ _{0}} \frac{q _{3}}{r _{3 P}} V = V_{1} + V_{2} + V_{3} + \dots\dots$

5.0Potential Due To A Point Charge

Derivation of expression for potential due to point charge Q, at a point which is at a distance r from the point charge.

From Definition of Potential

$V V V = \frac{W _{ε x t (\infty \to P)}}{q _{0}} = \frac{- \int _{\infty}^{r} ( q _{0} E ) \cdot d r}{q _{0}} = - \int_{\infty}^{r} E \cdot d r = - \int_{\infty}^{r} \frac{k Q}{r ^{2}} (- d r) cos 18 0^{\circ} = \frac{k Q}{r} = \frac{k Q}{r}$

Potential due to a Positive Point Charge

$V_{P} = + \frac{k Q}{r}$

Potential due to a Negative Point Charge

$V_{P} = - \frac{k Q}{r}$

Note: Where reference potential is 0 at infinity.

Illustration-1. Two point charges $2 μ C$ and $- 4 μ C$ are situated at points (-2 m,0 m) and (2 m,0 m) respectively. Find out potential at point C(4 m,0 m) and D $(0 m, 5 m)$ .

Solution:

Potential at Point C

$V_{C} = V_{q 1} + V_{q 2} = \frac{k ( 2 μ C )}{6} + \frac{k ( - 4 μ C )}{6} = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{6} - \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 6}}{2} = - 15000 V$

Similarly,

$V_{D} = V_{q 1} + V_{q 2} = \frac{k ( 2 μ C )}{( 5 ) ^{2} + 2 ^{2}} + \frac{k ( - 4 μ C )}{( 5 ) ^{2} + 2 ^{2}} = \frac{k ( 2 μ C )}{3} + \frac{k ( - 4 μ C )}{3} = - 6000 V$

Illustration-2.Find potential at origin O due to the given charge distributions.

(I)

(II)

Solution:

(i) $V_{0} = \frac{k q}{a} + \frac{k q}{2 a} + \frac{k q}{4 a} + \frac{k q}{8 a} + \dots\dots\dots\dots \infty = \frac{k q}{a} [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots .\infty] = \frac{k q}{a} [\frac{1}{1 - \frac{1}{2}}] \Rightarrow V_{0} = \frac{2 k q}{a}$

(ii) $V_{0} = \frac{k q}{a} - \frac{k q}{2 a} + \frac{k q}{4 a} - \frac{k q}{8 a} + \dots .\infty = \frac{k q}{a} [1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots .\infty] = \frac{k q}{a} [\frac{1}{1 + \frac{1}{2}}] \Rightarrow V_{0} = \frac{2 k q}{3 a}$

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Electric Potential in electromagnetism

Electric potential is a fundamental concept in electromagnetism that describes the amount of electric potential energy a unit charge possesses at a specific point in an electric field. It represents the work needed to move a positive test charge from a reference point, typically infinity, to that location without producing acceleration. Measured in volts (V), electric potential helps explain how charges interact, how electric circuits function, and how energy is transferred within electrical systems. Understanding electric potential is essential for studying electric fields, capacitors, and voltage in practical applications ranging from electronics to power distribution.

1.0Definition of Electric Potential

The potential energy per unit charge $U / q_{0}$ is independent of the value of $q_{0}$ and has a value at every point in an electric field. This quantity $U / q_{0}$ is called the electric potential (or simply the potential) V. Thus, the electric potential at any point in an electric field is:

$V = U / q_{0}$

Since potential energy is a scalar quantity so the electric potential also is a scalar quantity. If the test charge is moved between two positions A and B in an electric field, the charge-field system experiences a change in potential energy. Potential is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge $q_{0}$ .

V=Uq0=-ABE.dl

2.0Physical Interpretation of Electric Potential

In an electrostatic field, the electric potential (due to some source charges) at a point P is defined as the work done by an external agent in taking a unit point positive charge from a reference point (generally taken at infinity) to that point P without changing its kinetic energy.

$Δ V = \frac{Δ U}{q _{0}} = - \int_{A}^{B} E \cdot d l$

3.0Mathematical Representation of Electric Potential

If $(W_{\infty \to P})_{(e x t)}$ is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is

$V_{P} = [\frac{( W _{\infty \to P} ) _{e x t}}{q}]_{Δ K = 0} = \frac{( - W _{e l c} ) _{\infty \to P}}{q}$

Note:

(1) $(W_{\infty \to P})_{(e x t)}$ can also be called as the work done by external agents against the electric force on a unit positive charge due to the source charge.

(2) Write both W and q with proper signs.

4.0Properties of Electric Potential

Potential is a scalar quantity, its value may be positive, negative or zero.
SI unit of Potential is $volt = \frac{Joule}{Coulomb}$ and its dimensional formula is $[M^{1} L^{2} T^{- 3} I^{- 1}]$ .
Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point.
Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinity. $(Taking V_{\infty} = 0)$
Potential decreases in the direction of the electric field.
By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

$V_{1} = \frac{1}{4 π ϵ _{0}} \frac{q _{1}}{r _{1 P}}, V_{2} = \frac{1}{4 π ϵ _{0}} \frac{q _{2}}{r _{2 P}}, V_{3} = \frac{1}{4 π ϵ _{0}} \frac{q _{3}}{r _{3 P}} V = V_{1} + V_{2} + V_{3} + \dots\dots$

5.0Potential Due To A Point Charge

Derivation of expression for potential due to point charge Q, at a point which is at a distance r from the point charge.

From Definition of Potential

$V V V = \frac{W _{ε x t (\infty \to P)}}{q _{0}} = \frac{- \int _{\infty}^{r} ( q _{0} E ) \cdot d r}{q _{0}} = - \int_{\infty}^{r} E \cdot d r = - \int_{\infty}^{r} \frac{k Q}{r ^{2}} (- d r) cos 18 0^{\circ} = \frac{k Q}{r} = \frac{k Q}{r}$

Potential due to a Positive Point Charge

$V_{P} = + \frac{k Q}{r}$

Potential due to a Negative Point Charge

$V_{P} = - \frac{k Q}{r}$

Note: Where reference potential is 0 at infinity.

Illustration-1. Two point charges $2 μ C$ and $- 4 μ C$ are situated at points (-2 m,0 m) and (2 m,0 m) respectively. Find out potential at point C(4 m,0 m) and D $(0 m, 5 m)$ .

Solution:

Potential at Point C

$V_{C} = V_{q 1} + V_{q 2} = \frac{k ( 2 μ C )}{6} + \frac{k ( - 4 μ C )}{6} = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{6} - \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 6}}{2} = - 15000 V$

Similarly,

$V_{D} = V_{q 1} + V_{q 2} = \frac{k ( 2 μ C )}{( 5 ) ^{2} + 2 ^{2}} + \frac{k ( - 4 μ C )}{( 5 ) ^{2} + 2 ^{2}} = \frac{k ( 2 μ C )}{3} + \frac{k ( - 4 μ C )}{3} = - 6000 V$

Illustration-2.Find potential at origin O due to the given charge distributions.

(I)

(II)

Solution:

(i) $V_{0} = \frac{k q}{a} + \frac{k q}{2 a} + \frac{k q}{4 a} + \frac{k q}{8 a} + \dots\dots\dots\dots \infty = \frac{k q}{a} [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots .\infty] = \frac{k q}{a} [\frac{1}{1 - \frac{1}{2}}] \Rightarrow V_{0} = \frac{2 k q}{a}$

(ii) $V_{0} = \frac{k q}{a} - \frac{k q}{2 a} + \frac{k q}{4 a} - \frac{k q}{8 a} + \dots .\infty = \frac{k q}{a} [1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots .\infty] = \frac{k q}{a} [\frac{1}{1 + \frac{1}{2}}] \Rightarrow V_{0} = \frac{2 k q}{3 a}$

Frequently Asked Questions

What does electric potential at a point physically represent?

Why is electric potential a scalar quantity even though the electric field is a vector?

Why is the electric potential chosen to be zero at infinity?

Can two different charge configurations produce the same electric potential at a point?

What happens to the electric potential if we double all charges in a configuration?

Join ALLEN!

Electric Potential in electromagnetism

1.0Definition of Electric Potential

2.0Physical Interpretation of Electric Potential

3.0Mathematical Representation of Electric Potential

4.0Properties of Electric Potential

5.0Potential Due To A Point Charge

Table of Contents

Frequently Asked Questions

What does electric potential at a point physically represent?

Why is electric potential a scalar quantity even though the electric field is a vector?

Why is the electric potential chosen to be zero at infinity?

Can two different charge configurations produce the same electric potential at a point?

What happens to the electric potential if we double all charges in a configuration?

Join ALLEN!

Electric Potential in electromagnetism

1.0Definition of Electric Potential

2.0Physical Interpretation of Electric Potential

3.0Mathematical Representation of Electric Potential

4.0Properties of Electric Potential

5.0Potential Due To A Point Charge

Table of Contents