- THE ELECTROSTATIC POTENTIAL
- 25.1. Introduction
- 25.2. Calculating the Electrostatic Potential
- Example: Problem 25.21
- Example: Problem 25.15

- 25.3. The Electrostatic Field as a Conservative Field
- 25.4. The Gradient of the Electrostatic Potential
- Example: Problem 25.32
- Example: Problem 25.36

- 25.5. The Potential and Field of a Dipole

## 25.1. Introduction

The electrostatic force is a conservative force. This means that the work itdoes on a particle depends only on the initial and final position of theparticle, and not on the path followed. With each conservative force, apotential energy can be associated. The introduction of the potential energyis useful since it allows us to apply conservation of mechanical energy whichsimplifies the solution of a large number of problems.

The potential energy U associated with a conservative force F is defined inthe following manner

_{}

where U(P_{0}) is the potential energy at the reference positionP_{0} (usually U(P_{0}) = 0) and the path integral is along anyconvenient path connecting P_{0} and P_{1}. Since the force Fis conservative, the integral in eq.(25.1) will not depend on the path chosen.If the work W is positive (force and displacement pointing in the samedirection) the potential energy at P_{1} will be smaller than thepotential energy at P_{0}. If energy is conserved, a decrease in thepotential energy will result in an increase of the kinetic energy. If the workW is negative (force and displacement pointing in opposite directions) thepotential energy at P_{1} will be larger than the potential energy atP_{0}. If energy is conserved, an increase in the potential energywill result in an decrease of the kinetic energy. If In electrostatic problemsthe reference point P_{0} is usually chosen to correspond to aninfinite distance, and the potential energy at this reference point is taken tobe equal to zero. Equation (25.1) can then be rewritten as:

_{}

To describe the potential energy associated with a charge distribution theconcept of the ** electrostatic potential V** is introduced. Theelectrostatic potential V at a given position is defined as the potentialenergy of a test particle divided by the charge q of this object:

_{}

In the last step of eq.(25.3) we have assumed that the reference pointP_{0} is taken at infinity, and that the electrostatic potential atthat point is equal to 0. Since the force per unit charge is the electricfield (see Chapter 23), eq. (25.3) can be rewritten as

_{}

The unit of electrostatic potential is the ** volt** (V), and 1 V = 1J/C = 1 Nm/C. Equation (25.4) shows that as the unit of the electric fieldwe can also use V/m.

A common used unit for the energy of a particle is the electron-volt (eV)which is defined as the change in kinetic energy of an electron that travelsover a potential difference of 1 V. The electron-volt can be related to theJoule via eq.(25.3). Equation (25.3) shows that the change in energy of anelectron when it crosses over a 1 V potential difference is equal to 1.6^{.} 10^{-19} J and we thus conclude that 1 eV = 1.6^{.} 10^{-19} J

## 25.2. Calculating the Electrostatic Potential

A charge q is moved from P_{0} to P_{1} in the vicinity ofcharge q' (see Figure 25.1). The electrostatic potential at P_{1} canbe determined using eq. (25.4) and evaluating the integral along the path shownin Figure 25.1. Along the circular part of the path the electric field and thedisplacement are perpendicular, and the change in the electrostatic potentialwill be zero. Equation (25.4) can therefore be rewritten as

_{}

If the charge q' is positive, the potential increases with a decreasingdistance r. The electric field points away from a positive charge, and weconclude that the electric field points from regions with a high electrostaticpotential towards regions with a low electrostatic potential.

**Figure 25.1. Path followed by charge q between P**

_{0}andP_{1}._{}

### Example: Problem 25.21

A total charge Q is distributed uniformly along a straight rod oflength L. Find the potential at point P at a distance h from the midpoint ofthe rod (see Figure 25.2).

The potential at P due to a small segment of the rod, with length dx andcharge dQ, located at the position indicated in Figure 25.3 is given by

_{}

The charge dQ of the segment is related to the total charge Q and length L

_{}

Combining equations (25.7) and (25.8) we obtain the following expression fordV:

_{}

**Figure 25.2. Problem 25.21.**

**Figure 25.3. Solution of Problem 25.21.**

_{}

### Example: Problem 25.15

An alpha particle with a kinetic energy of 1.7 x 10^{-12} J isshot directly towards a platinum nucleus from a very large distance. What willbe the distance of closest approach ? The electric charge of the alphaparticle is 2e and that of the platinum nucleus is 78e. Treat the alphaparticle and the nucleus as spherical charge distributions and disregard themotion of the nucleus.

The initial mechanical energy is equal to the kinetic energy of the alphaparticle

_{}

Due to the electric repulsion between the alpha particle and the platinumnucleus, the alpha particle will slow down. At the distance of closestapproach the velocity of the alpha particle is zero, and thus its kineticenergy is equal to zero. The total mechanical energy at this point is equal tothe potential energy of the system

_{}

where q_{1} is the charge of the alpha particle, q_{2} is thecharge of the platinum nucleus, and d is the distance of closest approach.Applying conservation of mechanical energy we obtain

_{}

The distance of closest approach can be obtained from eq.(25.13)

_{}

## 25.3. The Electrostatic Field as a Conservative Field

The electric field is a conservative field since the electric force is aconservative force. This implies that the path integral

_{}

between point P_{0} and point P_{1} is independent of the pathbetween these two points. In this case the path integral for any closed pathwill be zero:

_{}

Equation (25.16) can be used to prove an interesting theorem:

" within a closed, empty cavity inside a homogeneous conductor, the electricfield is exactly zero ".

**Figure 25.4. Cross section of cavity inside sphericalconductor.**

## 25.4. The Gradient of the Electrostatic Potential

The electrostatic potential V is related to the electrostatic field E. If theelectric field E is known, the electrostatic potential V can be obtained usingeq.(25.4), and vice-versa. In this section we will discuss how the electricfield E can be obtained if the electrostatic potential is known.

**Figure 25.5. Calculation of the electric field E.**

_{1}and P

_{2}is given by

_{}

where the angle [theta] is the angle between the direction of the electricfield and the direction of the displacement (see Figure 25.5). Equation(25.17) can be rewritten as

_{}

where E_{L} indicates the component of the electric field along theL-axis. If the direction of the displacement is chosen to coincide with thex-axis, eq.(25.18) becomes

_{}

For the displacements along the y-axis and z-axis we obtain

_{}

_{}

The total electric field E can be obtained from the electrostatic potential Vby combining equations (25.19), (25.20), and (25.21):

_{}

Equation (25.22) is usually written in the following form

_{}

where --V is the gradient of the potential V.

In many electrostatic problems the electric field of a certain chargedistribution must be evaluated. The calculation of the electric field can becarried out using two different methods:

1.the electric field can be calculated by applying Coulomb's law and vectoraddition of the contributions from all charges of the charge distribution.

2.the total electrostatic potential V can be obtained from the algebraic sumof the potential due to all charges that make up the charge distribution, andsubsequently using eq.(25.23) to calculate the electric field E.

In many cases method 2 is simpler since the calculation of the electrostaticpotential involves an algebraic sum, while method 1 relies on the vector sum.

### Example: Problem 25.32

In some region of space, the electrostatic potential is the followingfunction of x, y, and z:

_{}

where the potential is measured in volts and the distances in meters. Find theelectric field at the points x = 2 m, y = 2 m.

The x, y and z components of the electric field E can be obtained from thegradient of the potential V (eq.(25.23)):

_{}

_{}

_{}

Evaluating equations (25.25), (25.26), and (25.27) at x = 2 m and y = 2 mgives

_{}

_{}

_{}

Thus

_{}

### Example: Problem 25.36

An annulus (a disk with a hole) made of paper has an outer radius R andan inner radius R/2 (see Figure 25.6). An amount Q of electric charge isuniformly distributed over the paper.

a)Find the potential as a function of the distance on the axis of theannulus.

b)Find the electric field on the axis of the annulus.

We define the x-axis to coincide with the axis of the annulus (see Figure25.7). The first step in the calculation of the total electrostatic potentialat point P due to the annulus is to calculate the electrostatic potential at Pdue to a small segment of the annulus. Consider a ring with radius r and widthdr as shown in Figure 25.7. The electrostatic potential dV at P generated bythis ring is given by

_{}

where dQ is the charge on the ring. The charge density [rho] of the annulus isequal to

_{}

_{}

Substituting eq.(25.34) into eq.(25.32) we obtain

_{}

The total electrostatic potential can be obtained by integrating eq.(25.35)over the whole annulus:

_{}

**Figure 25.7. Calculation of electrostatic potential in Problem25.36.**

_{}

Since the electrostatic field and the electrostatic potential are related wecan replace the field lines by so called ** equipotential surfaces**.Equipotential surfaces are defined as surfaces on which each point has the sameelectrostatic potential. The component of the electric field parallel to thissurface must be zero since the change in the potential between all points onthis surface is equal to zero. This implies that the direction of the electricfield is perpendicular to the equipotential surfaces.

## 25.5. The Potential and Field of a Dipole

Figure 25.8 shows an electric dipole located along the z-axis. It consists oftwo charges + Q and - Q, separated by a distance L. The electrostaticpotential at point P can be found by summing the potentials generated by eachof the two charges:

_{}

**Figure 25.8. The electric dipole.**

_{1}and r

_{2}are parallel.In this case

_{}

and

_{}

The electrostatic potential at P can now be rewritten as

_{}

where p is the dipole moment of the charge distribution. The electric field ofthe dipole can be obtained from eq.(25.41) by taking the gradient (seeeq.(25.23)).

Send comments, questions and/or suggestions via email to**wolfs@nsrl.rochester.edu**and/or visit the home page of Frank Wolfs.

## FAQs

### THE ELECTROSTATIC POTENTIAL? ›

The electric potential (also called the electric field potential, potential drop, the electrostatic potential) is defined as **the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field**.

### What is electrostatic potential answer? ›

The electrostatic potential, also known as the electric field potential, electric potential, or potential drop is defined as. **The amount of work done to move a unit charge from a reference point to a specific point inside the field without producing an acceleration**.

### What is electrostatic potential energy in physics? ›

Electric potential energy is **the energy that is needed to move a charge against an electric field**. You need more energy to move a charge further in the electric field, but also more energy to move it through a stronger electric field.

### What is the potential function in electrostatics? ›

The potential function of an electrostatic field is given by **V=2x2**.

### What is the formula of potential in electrostatics? ›

**V = k × [q/r]**

Where, V = electric potential energy. q = point charge. r = distance between any point around the charge to the point charge.