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
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 PotentialA charge q is moved from P0 to P1 in the vicinity of charge q' (see Figure 25.1). The electrostatic potential at P1 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
Example: Problem 25.21A 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
Example: Problem 25.15An 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
25.3. The Electrostatic Field as a Conservative FieldThe electric field is a conservative field since the electric force is a conservative force. This implies that the path integral
" within a closed, empty cavity inside a homogeneous conductor, the electric field is exactly zero ".
25.4. The Gradient of the Electrostatic PotentialThe 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.
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 25.32In 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 25.36An 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