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Electrostatic Potential Energy

  • Last Updated : 21 Jul, 2021

The electric potential at a place in an electric field is the amount of effort required to transport a unit positive charge from infinity to that point, whereas electric potential energy is the amount of energy required to move a charge against the electric field. The gravitational potential energy of a unit mass put at a certain position in the gravitational field is called gravitational potential. In this sense, the electric potential energy of a unit positive charge at any location in the electric field is equal to the electric potential energy stored in a unit positive charge at that point. The electric potential V at a given point in the field is given by,

 V = W/q

If W is the work done in transferring a unit positive charge q from infinity to a particular point in the field. It means that electric potential is measured in relation to some reference point and that we can only measure the difference in potential between two points, similar to potential energy. The scalar quantity is electric potential. A volt is a unit of measurement that is equivalent to joule per coulomb (J/C).

Potential energy Difference

Consider the electrostatic field E that exists as a result of a charge arrangement. Consider the field E caused by a charge Q placed at the origin for the sake of simplicity. Consider moving a test charge q from a point R to a point P while resisting the charge Q’s repulsive force. If Q and q are both positive or both negative, this will happen with reference. Let’s use Q as an example, with q > 0. 

A test charge q (> 0) is moved from the point R to the point P against the repulsive force on it by the charge Q (> 0)placed at the origin

Assume that the test charge q is so little that it has no effect on the original configuration, specifically the charge Q at the origin (or that Q is held fixed at the origin by some unknown force). Second, apply an external force Fext exactly enough to counter the repulsive electric force FE (i.e. Fext= –FE ) as the charge q  move from R to P. This means that when the charge q is transported from R to P, it experiences no net force or acceleration, implying that it is transported at an infinitesimally slow constant speed. In this case, the work done by the external force is the inverse that is minus the work done by the electric force, and the potential energy of the charge q is fully stored. If the external force is withdrawn when the charge reaches P, the electric force will pull the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q, preserving the sum of the kinetic and potential energies.



Therefore, the work done by external forces in moving a charge q from R to P can be written as,

W_{RP}=\int_{R}^{P} F_{ext}\cdot{dr}\\ W_{RP}=-\int_{R}^{P} F_{E}\cdot{dr}

The above expression is the work done against electrostatic opposing force and gets stored as potential energy.

A particle with charge q has a definite electrostatic potential energy at every location in the electric field, and the work done raises its potential energy by an amount equal to the potential energy difference between points R and P. Therefore, the potential energy difference can be expressed as,

∆U = UP –  UR = WRP

Note that this displacement is in the inverse direction of the electric force, hence the work done by the electric field is negative, i.e., –WRP.

As a result, the work required by an external force to move (without accelerating) charge q from one location to another for an electric field of any arbitrary charge configuration can be defined as the electric potential energy difference between two points.

At this point, two key points should be made,



  1. The work done by an electrostatic field in transferring a charge from one location to another is solely reliant on the initial and final points and is unaffected by the path used to get there. This is a conservative force’s defining attribute.
  2. The above expression defines the difference in potential energy in terms of a physically meaningful quantity of work. To within an additive constant, potential energy is clearly uncertain. This indicates that the actual value of potential energy has no physical significance; only the change in potential energy is important. We can always add an arbitrary constant to potential energy at any time since the potential energy difference will not change,

(UP – α ) – (UR – α ) = UP  -UR

To put it another way, the point where potential energy is zero can be chosen at will. Electrostatic potential energy 0 at infinity is a convenient choice. If we take the point R at infinity with this option,

W∞P = UP – U = UP

The above expression defines the potential energy of a charge q at any moment in time.

The work done by the external force (equal and opposite to the electric force) in bringing the charge q from infinity to that location (in the presence of field due to any charge configuration) is called potential energy of charge q at a point.

Electrostatic Potential

In general, think about any static charge configuration. A test charge’s potential energy q is defined in terms of the work done on it. This work is obviously proportional to q because the force at any position is qE, where E is the electric field at that site due to the given charge arrangement. As a result, dividing the work by the charge q yields a quantity that is independent of q. To put it another way, the amount of work done per unit test charge represents the electric field associated with the charge arrangement. This leads to the concept of electrostatic potential V as a result of a particular charge arrangement.

W=V_P-V_R\\ W= \frac{U_P-U_R}{q}

where, VP and VR are the electrostatic potentials at P and R, respectively.

The work done by an external force to carry a unit positive charge from infinity to a location is equal to the electrostatic potential (V) at that point.

In other words, the electrostatic potential (V ) at any place in an area with an electrostatic field is the work needed to transport a unit positive charge from infinity to that location (without acceleration). Electric potential energy is a scalar quantity with no direction and only magnitude. It is symbolized by V and has the dimensional formula ML2T-3A-1.



Work done on a test charge q by the electrostatic field due to any given charge configuration is independent of the path and depends only on its initial and final positions.

The same caveats that were expressed before about potential energy apply to the definition of potential. To calculate the work done per unit test charge, start with an infinitesimal test charge δq, calculate the work done δw in bringing it from infinity to the point, and divide by δq to get the δW/δq ratio. In addition, at each point along the path, the external force must be equal to and opposite to the electrostatic force acting on the test charge.

Electric Potential of a Point Charge

Work done in bringing a unit positive test charge from infinity to the point P, against the repulsive force of charge Q (Q > 0), is the potential at P due to the charge Q.

Take a look at the origin of a point charge Q. Consider Q to be a positive character. We wish to find the potential at any location P using the position vector r from the origin. To do so, we need to figure out how much work it takes to transfer a unit positive test charge from infinity to point P. When Q > 0, the work done against the repulsive force on the test charge is positive. Because the work is independent of the path, we choose a convenient path — along the radial direction from infinity to point P.

The electrostatic force on a unit positive charge at some intermediate point P′ on the path equals to

\frac{Q\times1}{4\pi\epsilon_0r'^2}\hat{r'}

where  \hat{r'}     is the unit vector along OP′ therefore, work done against this force from r′ to r′ + ∆r′ can be written as

\Delta{W}=-\frac{Q}{4\pi\epsilon_0r'^2}\Delta{r'}

The negative sign represents ∆r′ < 0, ∆W is positive . Total work done (W) by the external force is determine by integrating the above equation both side, from r′ = ∞ to r′ = r,

W=-\int_{∞}^{r} \frac{Q}{4\pi\epsilon_0r'^2}d{r'}\\ W=\left[\frac{Q}{4\pi\epsilon_0r'}\right]_∞^r\\ W=\frac{Q}{4\pi\epsilon_0r}

The potential at P due to the charge Q can be expressed as

V(r)=\frac{Q}{4\pi\epsilon_0r}



Potential Energy of a System of Charges

Potential at a point due to a system of charges is the sum of potentials due to individual charges.

Suppose a system of charges q1 , q2 ,…, qn with position vectors r1 , r2 ,…, rn relative to some origin. The potential V1 at P due to the charge q1 can be expressed as

V_1=\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1P}}

where r1P is the distance between q1 and P.

Similarly, the potential V2 at P due to q2 and V3 due to q3 can be written as,

V_2=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r_{2P}}\\ V_3=\frac{1}{4\pi\epsilon_0}\frac{q_3}{r_{3P}}

where r2P and r3P are the distances of P from charges q2  and q3, respectively; and so on for the potential due to other charges. 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 charge that is,

V = V1 + V2 + V3 +…… + Vn

The above expression can be expressed as,

V=\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1P}}+\frac{1}{4\pi\epsilon_0}\frac{q_2}{r_{2P}}+ \frac{1}{4\pi\epsilon_0}\frac{q_3}{r_{3P}}+.....+\frac{1}{4\pi\epsilon_0}\frac{q_n}{r_{nP}}\\ V=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}+\frac{q_3}{r_{3P}}+...+\frac{q_n}{r_{nP}}\right)

It is necessary to divide a continuous charge distribution with a charge density (r) into small volume elements of size ∆v, each carrying a charge ρ∆v. Then, for each volume element, compute the potential and add (or, more properly, integrate) all of these contributions to get the overall potential owing to the distribution.



Sample Problems  

Problem 1: Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge and a 12.0 V car battery that can move 60,000 C of charge. How much energy does each deliver? 

Solution:

Given,

The value of the battery is 12.0 V battery that is its terminals have a 12.0 V potential difference.

The charge that the motorcycle battery move is 5000 C.

The 12.0 V car battery can move 60,000 C of charge.

When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to ΔPE = qΔV.

For the motorcycle battery, q = 5000 C and ΔV = 12.0 V. The total energy delivered by the motorcycle battery is

ΔPEmotorcycle = (5000 C) × (12.0 V)

ΔPEmotorcycle = 6.00 × 104  J

Now, for the car battery,

ΔPEcar = (60,000 C) × (12.0 V)

ΔPEcar =7.20 × 105 J

Problem 2: 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:

Given,

The mass of the particle is 40 mg.

The charge of the particle is 5×10-9 C.

The fixed positive point charge of magnitude 10-8 C.

The particle comes to rest momentarily at a distance r from the fixed charge, from conservation of energy we have,

According to the law of conservation of energy, the total energy of the system = Constant

 i.e. (K.E + P.E) = constant.

The expression for the kinetic energy can be expressed as,

K.E= \frac{1}{2}mu^2    

The expression for the potential energy can be expressed as,

Now,

(1/2)mu2 + (1/4)πεo × [Qq/a] = (1/4)πεo × [Qq/r]

Substituting the values in the above equation,

1/2 × 40 × 10-6 × 1/2 × 1/2 = 9 × 109 × 10-8 × 5 × 10-9 × [ 1/r – 1/(10 × 10-2)]

or, [1/r – 10] = (5×10-5)/(9×5×10-8) = 100/9



or, 1/r = (100/9) + 10 

or, 1/r  = 190/9 m

or r = 4.7 × 10-2 m

Since, F = [1/4πεo] × [Qq/r2]

Therefore, acceleration = F/m ∝ 1/r2 i.e., acceleration is not constant during motion.

Problem 3: 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. What is the velocity of the ball at point A, if at point B, it is 25 cm per second?

Solution:

Given,

The mass ball is  5 g.

The charge of the particle is 10-7 C.

The potential of ball at point A is 500 V and potential at point B is zero.

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

The work done on the charge by the field given by,

W = q × (VA – VB

Substitute the value in the above expression,

W = 10-7 × (500 – 0) 

W = 5 × 10-5 J

Therefore, 

W = (1/2) mv2 – (1/2) mu2

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

2 × 10-2 = 1/16 – u2

u2 = (1/16) – 0.02

u = (1- 0.32)/16

u = 0.0425

Therefore, u =0.206 m/s 

u = 20.6 cm/sec.

Problem 4: When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second?

Solution:

The expression for the potential energy can be written as,

ΔPE = qΔV

Rearrange the above expression,

q = ΔPE/ΔV

Substitute the values in the above equation,

q = −30.0 J/ 12.0 V

q = −30.0 J/ 12.0 J/C 

q = −2.50 C

The number of electrons n can be calculated as,

 n = q/e

 n = −2.50 C/(−1.60 ×10−19 C/e)

n = 1.56×1019 electrons

Problem 5: 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:

Given,

Two charges of magnitude 3C and 5C.

The separation between two charges are 0.5 m.

The potential at P due to the charge Q can be expressed as

 E_g =\frac{Q}{4\pi\epsilon_0r}

∆E = E0 – Eg

∆E = 0 – [-(9 × 109 × 5 × 3)/0.5] = 27 × 1010.

Therefore, ∆E = 27 × 1010 V.

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