The magnitude of electric potential energy possessed by an object is determined by two elements: the charge of the object itself and the relative position of the object with respect to other electrically charged objects. This potential energy is calculated by the amount of work done in moving the object from one point to another against the electric field.

The electric potential of any charge is calculated by dividing the electric potential energy gained when an object is moved against the electric field by the quantity of charge.

Table of Contents

• Electric Potential Definition
• Electric Potential Formula
• Electric Potential Derivation
• Electric Potential of one Point Charge
• Electric Potential for Multiple Charges
• Important Points About Electric Potential
• Solved Examples on Electric Potential

What is Electric Potential Energy?

Electric Potential Energy is the energy stored in an electric field due to the position of charged particles. It is the energy that a charged particle has due to its position relative to other charges.

The electric potential energy of any given charge or system of charges is referred to as the total work done by an external agent in bringing the charge or the system of charges from infinity to its current configuration without experiencing any acceleration.

Definition: Electric potential energy is the energy possessed by a unit charge due to its position in an electric field.

Overview

Electric potential energy is a scalar quantity and possesses only magnitude and no direction. It is measured in terms of Joules and is denoted by V. It has the dimensional formula of ML²T^-3A^-1.

Electric Potential
Denoted by	V, ∆V, U, ∆U
Dimension:	ML2T-3A-1
General Formula	Voltage = Energy / Charge
SI Unit	Volt

There are two factors that affect the electric potential energy of an object:

• Mass of the object
• Electric charge of the object

It has its own electric charge.

Its position in relation to other electrically charged objects.

Also Read:

• Electrostatics
• Equipotential Surface

Electric Potential Formula:

$$V = \frac{kQ}{r}$$

The electric potential energy of the system consisting of two charges, q₁ and q₂, separated by a distance d, is measured by the work done in moving the charge from infinity to that point against the electric field.

$$U = \frac{q_1q_2}{4\pi\epsilon_0 d}$$

If two like charges (two protons or two electrons) are brought towards each other, the potential energy of the system increases. If two unlike charges i.e. a proton and an electron are brought towards each other, the electric potential energy of the system decreases.

Electric Potential Formula

Method 2:

The electric potential at any point around a point charge q is given by:

$V=k\times \frac{q}{r}$

where

• V = Electric Potential Energy
• Q = Point Charge
• r = distance between any point around the charge and the point charge
• k = Coulomb constant; k = 9.0× 10⁹ N

Using Coulomb’s Law

The electrostatic potential between any two arbitrary charges q₁, q₂ separated by distance r is given by Coulomb’s law and mathematically written as:

$U=k\times \frac{q_1{q}_2}{r^2}$

The electrostatic potential energy is represented by U.

Q₁ and Q₂ are the two charges.

Note: The electric potential when $r=\mathrm{\infty }$ is equal to zero (as, $r=\mathrm{\infty }$ in the above formula).

Electric Potential Derivation

The total electric potential of two charges, q₁ and q₂, placed at a distance r from each other is defined as the total work done by an external force in bringing the charge from infinity to the given point.

-We can write it as, $$-\int_{r_a}^{r_b} F \mathrm{d}r = -(U_a - U_b)$$

Here, we see that the point r is present at infinity and the point r_a is r.

$-\int (r\to \infty )Fdr=-(U_r-U_{\infty })$

As we know, Unity is equal to zero.

Therefore, ${\int }_{r\to \mathrm{\infty }}F\cdot dr=-UR$

Using Coulomb’s law, the force between the two charges can be written as:

⇒ ${\int }_{r\to \mathrm{\infty }}\frac{-k{q}^2}{r^2}dr=-UR$

Or, $$-k \times \frac{qqo}{r} = UR$$

Therefore, UR = -kqqo/r

Electric Potential of a Point Charge

Let us consider a point charge q in the presence of another charge Q with infinite separation between them.

$UE(r)=ke\times \frac{qQ}{r}$

Let $k_e=\frac{1}{4\pi {ϵ}_0}=$ Columb’s constant

Let us consider a point charge q in the presence of several point charges Qi with infinite separation between them.

$UE(r)=keq\times \sum _{i=1}^n\frac{Q_i}{r_i}$

Electric Potential for Multiple Charges

In the Case of 3 Charges:

If three charges q1, q2 and q3 are situated at the vertices of a triangle, the potential energy of the system is given by $$U=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_3q_1}{r_{31}}\right)$$.

In the Case of 4 Charges:

The electric potential energy of the system, if four charges q1, q2, q3, and q4 are situated at the corners of a square, is:

$U=(1/4\pi \epsilon o)\times [(q1q2/d)+(q2q3/d)+(q3q4/d)+(q4q1/d)+(q4q2/\surd 2d)+(q3q1/\surd 2d)]$

Special Case:

If a charge q is moved against the electric field from a distance a to a distance b from a charge Q, the work done is given by:

$W=(V_b-V_a)\times q=[\frac{1}{4\pi {ϵ}_o}\times \left(\frac{Q_q}{b}\right)]-[\frac{1}{4\pi {ϵ}_o}\times \left(\frac{Q_q}{a}\right)]=\frac{Q_q}{4\pi {ϵ}_o}[\frac{1}{b}-\frac{1}{a}]=\frac{Q_q}{4\pi {ϵ}_o}\left[\frac{a-b}{ab}\right]$

Important Points

At a point midway between two equal and opposite charges, the electric potential is zero but the electric field is not necessarily zero.

The electric potential at a point is said to be one volt if one joule of work is done in moving one Coloumb of the charge against the electric field.

If a negative charge is moved from point A to B, the electric potential of the system will increase.

The reference level used to define electric potential at a point is infinity, signifying that the force on a test charge is zero at the reference level.

The surface of the Earth is taken to be at zero potential, as the Earth is so large that the addition or removal of charge from it will not affect its electrical state.

⇒ Check: Gauss’s Law and Its Applications

What is Electric Potential Difference?

Electric Potential Difference (also known as Voltage) is the difference in electric potential energy between two points in a circuit. It is measured in volts (V).

The potential between two points (E) in an electrical circuit is equal to the work done (W) by an external agent in moving a unit charge (Q) from one point to another.

Mathematically we can say that,

W = E*Q

• E = Electrical Potential Difference between Two Points
• W = Work done in moving a charge from one point to another
• Q = Quantity of charge in coulombs

Solved Examples on Electric Potential

Problem 1:

A particle of mass 40 mg and carrying a charge 5×10^-9 C is moving directly towards a fixed positive point charge of magnitude 10^-8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during motion?

Solution:

If the particle comes to rest momentarily at a distance r from the fixed charge, from conservation of energy, we can conclude that:

Total Energy of the system = Constant

(K.E + P.E) = Constant

$(1/2){\mu }_2+(1/4\pi {ϵ}_0)\times [Qq/a]=(1/4\pi {ϵ}_0)\times [Qq/r]$

Substituting the given data, we get:

$1/2\times 40\times 10⁻⁶\times 1/2\times 1/2=9\times 10⁹\times 10⁻⁸\times 5\times 10⁻⁹\times [1/r-1/(10\times 10⁻²)]$

$[1/r]=(5\times 10⁻⁵)/(9\times 5\times 10⁻⁸)=100/9$

$1/r=(190/9)m$

i.e., r = 4.7 × 10^-2 m

Since, F = $$\frac{1}{4\pi\epsilon_0}\times\frac{Qq}{r^2}$$

Therefore, acceleration = $$\frac{F}{m} \propto \frac{1}{r^2}$$, i.e., acceleration is not constant during motion.

Problem 2:

What is the velocity of the ball at point A, if at point B, it is 25 cm per second, given that a ball of mass 5 g and charge 10^-7 C moves from point A whose potential is 500 V to a point B whose potential is zero? Velocity of the ball

Solution:

Let u be the velocity of the ball at point A.

Work done on the charge by the field

$W=q\times (V_A–V_B)={10}^{-7}\mathrm{}\times (500–0)=5\times {10}^{-5}\mathrm{}J$

This appears in front of the increased Kinetic Energy.

W = (1/2)mv² - (1/2)mu²

5 × 10^-5 = (1/2) × 5/1000 [(1/4)² – u²]

$2\times {10}^{-}2=1/16-u^2$

$u2=\frac{1}{16}-0.02$

$\frac{1-0.32}{16}$

0.0425

Therefore, u = 0.206 m/s = 206 cm/min

Therefore, u = 20.6 cm/s.

Example 3: Let us say we have two charges of magnitude 1C and 2C placed at a distance 2 metre from each other. Calculate the electric potential between these two charges. (Take: k = 1)

Answer:

Given that, the magnitudes of charges are q1 = 1C and q2 = 2C.

The distance between these two charges is r = 2m.

The electric potential between these two charges is given by, $$U_r = -\frac{kq_o}{r}$$

Substituting the given values in the above equation yields:

Ur = -1 J.

Example 4:

How much work is required to be done, in order to bring two charges of magnitude 3C and 5C from a separation of infinite distance to a separation of 0.5 m?

Solution:

$\Delta E=E₀-Eₐ$

$=0-[-(9\times {10}^9\times 5\times 3)/0.5]=27\times {10}^{10}.$

Therefore, ΔE = 27 x 10¹⁰.