# 25.1. Introduction

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

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

_{ }

_{0}) is the potential energy at the reference position P

_{0}(usually U(P

_{0}) = 0) and the path integral is along any convenient path connecting P

_{0}and P

_{1}. Since the force F is 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 same direction) the potential energy at P

_{1}will be smaller than the potential energy at P

_{0}. If energy is conserved, a decrease in the potential energy will result in an increase of the kinetic energy. If the work W is negative (force and displacement pointing in opposite directions) the potential energy at P

_{1}will be larger than the potential energy at P

_{0}. If energy is conserved, an increase in the potential energy will result in an decrease of the kinetic energy. If In electrostatic problems the reference point P

_{0}is usually chosen to correspond to an infinite distance, and the potential energy at this reference point is taken to be equal to zero. Equation (25.1) can then be rewritten as:

_{ }

**is introduced. The electrostatic potential V at a given position is defined as the potential energy of a test particle divided by the charge q of this object:**

__electrostatic potential V___{ }

_{0}is taken at infinity, and that the electrostatic potential at that point is equal to 0. Since the force per unit charge is the electric field (see Chapter 23), eq. (25.3) can be rewritten as

_{ }

**(V), and 1 V = 1 J/C = 1 Nm/C. Equation (25.4) shows that as the unit of the electric field we can also use V/m.**

__volt__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 travels over a potential difference of 1 V. The electron-volt can be related to the Joule via eq.(25.3). Equation (25.3) shows that the change in energy of an electron 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 of charge q' (see Figure 25.1). The electrostatic potential at P

_{1}can be determined using eq. (25.4) and evaluating the integral along the path shown in Figure 25.1. Along the circular part of the path the electric field and the displacement are perpendicular, and the change in the electrostatic potential will be zero. Equation (25.4) can therefore be rewritten as

_{ }

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

_{0}and P_{1}._{ }

### Example: Problem 25.21

A total charge Q is distributed uniformly along a straight rod of length L. Find the potential at point P at a distance h from the midpoint of the rod (see Figure 25.2).The potential at P due to a small segment of the rod, with length dx and charge dQ, located at the position indicated in Figure 25.3 is given by

_{ }

_{ }

_{ }

**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 is shot directly towards a platinum nucleus from a very large distance. What will be the distance of closest approach ? The electric charge of the alpha particle is 2e and that of the platinum nucleus is 78e. Treat the alpha particle and the nucleus as spherical charge distributions and disregard the motion of the nucleus.

The initial mechanical energy is equal to the kinetic energy of the alpha particle

_{ }

_{ }

_{1}is the charge of the alpha particle, q

_{2}is the charge of the platinum nucleus, and d is the distance of closest approach. Applying conservation of mechanical energy we obtain

_{ }

_{ }

## 25.3. The Electrostatic Field as a Conservative Field

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

_{ }

_{0}and point P

_{1}is independent of the path between these two points. In this case the path integral for any closed path will be zero:

_{ }

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

**Figure 25.4. Cross section of cavity inside spherical conductor.**

## 25.4. The Gradient of the Electrostatic Potential

**The electrostatic potential V is related to the electrostatic field E. If the electric field E is known, the electrostatic potential V can be obtained using eq.(25.4), and vice-versa. In this section we will discuss how the electric field 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

_{ }

_{ }

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

_{ }

_{ }

_{ }

_{ }

_{ }

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

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

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

In many cases method 2 is simpler since the calculation of the electrostatic potential 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 following function of x, y, and z:_{ }

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

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

### Example: Problem 25.36

An annulus (a disk with a hole) made of paper has an outer radius R and an inner radius R/2 (see Figure 25.6). An amount Q of electric charge is uniformly distributed over the paper.a) Find the potential as a function of the distance on the axis of the annulus.

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 Figure 25.7). The first step in the calculation of the total electrostatic potential at point P due to the annulus is to calculate the electrostatic potential at P due to a small segment of the annulus. Consider a ring with radius r and width dr as shown in Figure 25.7. The electrostatic potential dV at P generated by this ring is given by

_{ }

_{ }

**Figure 25.6. Problem 25.36.**

_{ }

_{ }

_{ }

**Figure 25.7. Calculation of electrostatic potential in Problem 25.36.**

_{ }

**. Equipotential surfaces are defined as surfaces on which each point has the same electrostatic potential. The component of the electric field parallel to this surface must be zero since the change in the potential between all points on this surface is equal to zero. This implies that the direction of the electric field is perpendicular to the equipotential surfaces.**

__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 of two charges + Q and - Q, separated by a distance L. The electrostatic potential at point P can be found by summing the potentials generated by each of the two charges:**

_{ }

**Figure 25.8. The electric dipole.**

_{1}and r

_{2}are parallel. In this case

_{ }

_{ }

_{ }

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