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# electric potential energy point charge

From Wikipedia

Electric potential energy

Electric potential energy, or electrostatic potential energy, is a potential energy associated with the conservativeCoulomb forces within a defined system of point charges. The term "electrostatic potential energy" is preferred here because it seems less likely to be misunderstood. The reference zero is usually taken to be a state in which the individual point charges are very well separated ("are at infinite separation") and are at rest. The electrostatic potential energy of the system (UE), relative to this zero, is equal to the total workW that must be done by a hypothetical external agent in order to bring the charges slowly, one by one, from infinite separation to the desired system configuration:

U_{\mathrm{E}} = \; W \;.

In this process the external agent is deemed to provide or absorb any relevant work, and the point charge being slowly moved gains no kinetic energy.

Sometimes reference is made to the potential energy of a charge in an electrostatic field. This actually refers to the potential energy of the system containing the charge and the other charges that created the electrostatic field.

To calculate the work required to bring a point charge into the vicinity of other (stationary) point charges, it is sufficient to know only (a) the total field generated by the other charges and (b) the charge of the point charge being moved. The field due to the charge being moved and the values of the other charges are not required. Nonetheless, in many circumstances it is mathematically easier to add up all the pairwise potential energies (as below).

It is important to understand that electrostatics is a 18th-19th-century theory of hypothetical entities called "point charges". Electrostatics is categorically not a complete theory of the charged physical particles that make up the physical world, and which are subject to the Heisenberg uncertainty principle and other laws of quantum mechanics.

## Electrostatic potential energy stored in a configuration of discrete point charges

The mutual electrostatic potential energy of two charges is equal to the potential energy of a charge in the electrostatic potential generated by the other. That is to say, if charge q_1 generates an electrostatic potential \Phi_1(\mathbf r), which is a function of position \mathbf r, then U_E = q_2 \Phi_1(\mathbf r_2). Also, a similar development gives U_E = q_1 \Phi_2(\mathbf r_1) This can be generalized to give an expression for a group of N charges, q_i at positions \mathbf r_i:

U_E = \frac{1}{2}\sum_i^N q_i \Phi(\mathbf r_i)

where, for each i value, \Phi(\mathbf r_i) is the electrostatic potential due to all point charges except the one at \mathbf r_i

Note: The factor of one half accounts for the 'double counting' of charge pairs. For example, consider the case of just two charges.

Alternatively, the factor of one half may be dropped if the sum is only performed once per charge pair. This is done in the examples below to cut down on the math.

#### One point charge

The electrostatic potential energy of a system containing only one point charge is zero, as there are no other sources of electrostatic potential against which an external agent must do work in moving the point charge from infinity to its final location. One should carefully consider the possibility of the point charge interacting with its own electrostatic potential. However, since such a potential at the location of the point charge itself is infinite, this "self-energy" is intentionally excluded from an evaluation of the total (finite) electrostatic potential energy of the system. Moreover, one may argue that since the electrostatic potential due to the point charge itself provides no work in moving the point charge around this interaction is unimportant for most purposes.

#### Two point charges

Consider bringing a second point charge, q_2 , into its final position in the vicinity of the first point charge, q_1. The electrostatic potential \Phi(r) due to q_1 is

\Phi(r_1) = \; k_{\mathrm{e}} \frac{q_1}{r}

where ke is Coulomb's constant. In the International System of Quantities, which has been the preferred international system since the 1970s and forms the basis for the definition of SI units, the Coulomb constant is given by

k_{\mathrm{e}} = \; \frac{1}{4 \pi \varepsilon_0} ,

where \varepsilon_0 is the electric constant. Hence we obtain,

U_\mathrm{E} = \; \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{ r_{12}}

where r_{12} is the distance between the two point charges.

The electrostatic potential energy is negative if the charges have opposite sign and positive if the charges have the same sign. Negative mutual potential energy corresponds to attraction between two point charges; positive mutual potential energy to repulsion between two point charges.

### Three or more point charges

For three or more point charges, the electrostatic potential energy of the system may be calculated by the total amount of work done by an external agent in bringing individual point charges into their final positions one after another. Thus,

U_\mathrm{E} = \frac{1}{4 \pi \varepsilon_0} \left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} + \cdots + \frac{q_i q_j}{r_{ij}} \right)

where

q1, q2, ..., are the point charges
rijis the distance between the ith and jth point charges.

NOTE Here, \varepsilon_0 is the relative permittivity of free space. When the charge is in a medium other than free space / air ,the relative permittivity, \varepsilon = k\varepsilon_0, has to be taken into account where k is the dielectric constant of the medium. K is the ratio of the electrostatic force on the charges in free space to the electrostatic force on the charges in the respective medium

## Energy stored in an electrostatic field distribution

One may take the equation for the electrostatic potential energy of a continuous charge distribution and put it in terms of the electrostatic field.

Since Gauss' law for electrostatic field in differential form states

\mathbf{\nabla}\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}

where

• \mathbf{E} \ is the electric field vector
• \rho \ is the total charge density including dipole charges

Question:Three point charges Q1 = 2.6 C, Q2 = 5.4 C, and Q3 = -3.9 C are arranged as shown in the figure below. What is the total electric potential energy of this system? heres the link to the image: http://i965.photobucket.com/albums/ae135/Peal0708/18-p-009.gif

Answers:Three or more point charges For 3 or more point charges, the electrostatic potential energy of the system may be calculated by bringing individual charges into position one after another, and taking the sum of the energies required to bring each additional charge into position. Thus U=1/4pi*epsilon (q1*q2/r12 + q1*q3/r13 + q3*q2/r32) solve :)

Question:Three point charges Q1 = 2.6 C, Q2 = 5.4 C, and Q3 = -3.9 C are arranged as shown in the figure below. What is the total electric potential energy of this system? answer in J please and thanks! heres the link to the image: http://i965.photobucket.com/albums/ae135/Peal0708/18-p-009.gif

Answers:U = k*(q1*q2/r12 +q1*q3/r13 + q2*q3/r23) = 9.0x10^9*(2.6x10^-6*5.4x10^-6)/sqrt(1.5^2+1.5^2) + 2.6x10^-6*(-3.9x10^-6)/sqrt(1.5^2+3^2) + 5.4x10^-6*(-3.9x10^-6)/4.5) = -9.76x10^-3J

Question:A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy of the pair of charges is +5.4x10^-8J. When the second charge is moved to point , the electric force on the charge does -1.9x10^-8J of work.

Answers:____________________________________ A point charge q1 is held stationary at the origin. A second charge q2 is placed at point 'a', and the electric potential energy of the pair of charges is +5.4x10^-8J. The electric potential energy of the pair of charges =PE=kq1q2/a= +5.4x10^-8J. When the second charge is moved to point , the electric force on the charge does -1.9x10^-8J of work. Gain in PE =1.9x10^-8J Final PE at 'b'=kq1q2/b= [5.4x10^-8 +1.9x10^-8]=7.3*10^-8 J The electric potential energy of the pair of charges when the second charge is at point 'b' is 7.3*10^- 8 J _____________________________

Question:What are the two charges?

Answers:We know q1+q2 = 3E-8 C and we know q1 and q2 are of opposite polarity. PE = kq1q2/r ==> q1q2 = PE*r/k = -4.1666667E-10 C Then q1 + (q1q2)/q1 = (q1+q2) ==> q1^2 - (q1+q2)q + (q1q2) = 0 This is a quadratic with a = 1, b = -3E-8 and c = -4.1666667E-10. The positive root is q1 = 2.042742E-5 C With this value for q1, q2 = -q1+3E-8 C = -2.039742E-5 C, which not surprisingly is the negative root of the quadratic. Obviously you can swap these values between q1 and q2. P.S. I would appreciate it if you didn't delete this Q&A as you have done with a previous question I answered, since I often use my old work as a basis for new answers. Thanks.