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Electric Potential
Equipotential Surfaces; Potential due to a Point Charge and a Group of Point Charges; Potential due to an Electric Dipole; Potential due to a Charge Distribution; Relation between Electric Field and Electric Potential Energy
- How to find the electric force on a particle 1 of charge q 1 when the particle is placed near a particle 2 of charge q 2.
- A nagging question remains: How does particle 1 “know “of the presence of particle 2?
- That is, since the particles do not touch, how can particle 2 push on particle 1—how can there be such an action at a distance?
- One purpose of physics is to record observations about our world, such as the magnitude and direction of the push on particle 1. Another purpose is to provide a deeper explanation of what is recorded.
- One purpose of this chapter is to provide such a deeper explanation to our nagging questions about electric force at a distance. We can answer those questions by saying that particle 2 sets up an electric field in the space surrounding itself. If we place particle 1 at any given point in that space, the particle “knows” of the presence of particle 2 because it is affected by the electric field that particle 2 has already set up at that point. Thus, particle 2 pushes on particle 1 not by touching it but by means of the electric field produced by particle 2. One purpose of physics is to record observations about our world,
Equipotential Surfaces:
Potential energy:
When an electrostatic force acts between two or more charged particles within a system of particles, we can assign an electric potential energy U to the system. If the system changes its configuration from an initial state i to a different final state f, the electrostatic force does work W on the particles. The change ΔU in the potential energy of the system is
. ΔU = Uf - Ui = -W (1) The work done by the electrostatic force is path independent. The work W done by the force on the particle is the same for all paths between points i and f.
Also, we usually set the corresponding reference potential energy to be zero. Suppose that several charged particles come together from initially infinite separations (state i) to form a system of neighboring particles (state f ). Let the initial potential energy Ui be zero, and let Wf represent the work done by the electrostatic forces between the particles during the move in from infinity. Then, the final potential energy U of the system is U = - W∞
Electric Potential:
An electric potential (also called the electric field potential , potential drop or the electrostatic potential ) is the amount of work needed to move a unit of positive charge from a reference point to a specific point inside the field without producing acceleration. The electric potential , or voltage , is the difference in potential energy per unit charge between two locations in an electric field. Electric potential , the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. Typically, the reference point is Earth, although any point beyond the influence of the electric field charge can be used. The potential energy of a charged particle in an electric field depends on the charge magnitude.
The potential difference between two points is thus the negative of the work done by the electrostatic force to move a unit charge from one point to the other. A potential difference can be positive, negative, or zero, depending on the signs and magnitudes of q and W. If we set Ui = 0 at infinity as our reference potential energy, then by Eq. 1, the electric potential V must also be zero there. Then from Eq. 3, we can define the electric potential at any point in an electric field to be
Work Done by an Applied Force
Suppose we move a particle of charge q from point i to point f in an electric field by applying a force to it. During the move, our applied force does work W app on the charge while the electric field does work W on it. By the work - kinetic energy theorem, the change ΔK in the kinetic energy of the particle is
(1)
Now suppose the particle is stationary before and after the move. Then Kf and Ki are both zero, and Eq. 1 reduces to Wapp = - W. (2) The work Wapp done by our applied force during the move is equal to the negative of the work W done by the electric field - provided there is no change in kinetic energy. we can relate the work done by our applied force to the change in the potential energy of the particle during the move. As we know that ΔU = Uf - Ui = -W, so
ΔU = Uf - Ui = Wapp (3) We can relate our work Wapp to the electric potential difference Δ V between the initial and final locations of the particle.
Equipotential Surfaces Adjacent points that have the same electric potential form an equipotential surface, which can be either an imaginary surface or a real, physical surface. No net work W is done on a charged particle by an electric field when the particle moves between two points i and f on the same equipotential surface. W must be zero if Vf = Vi.
Figure 1 below shows a family of equipotential surfaces associated with the electric field due to some distribution of charges. The work done by the electric field on a charged particle as the particle moves from one end to the other of paths I and II is zero because each of these paths begins and ends on the same equipotential surface and thus there is no net change in potential. The work done as the charged particle moves from one end to the other of paths III and IV is not zero but has the same value for both these paths because the initial and final potentials are identical for the two paths; that is, paths III and IV connect the same pair of equipotential surfaces.
Electric field lines (purple) and cross sections of equipotential surfaces (gold) for ( a ) a uniform electric field, ( b ) the field due to a point charge, and ( c ) the field due to an electric dipole.
Calculating the Potential from the Field
We can calculate the potential difference between any two points i and f in an electric field if we know the electric field vector E all along any path connecting those points. To make the calculation, we find the work done on a positive test charge by the field as the charge moves from i to f.
Consider an arbitrary electric field, represented by the field lines in Fig. above, and a positive test charge q 0 that moves along the path shown from point i to point f. At any point on the path, an electrostatic force q 0 E acts on the charge as it moves through a differential displacement. We know that the differential work dW done on a particle by a force F during a displacement ds is given by the dot product of the force and the displacement:
To find the total work W done on the particle by the field as the particle moves from point i to point f , the differential works done on the charge as it moves through all the displacements along the path:
Next, we set Vf = 0 (at ∞) and Vi = V (at R). Then, for the magnitude of the electric field at the site of the test charge, we substitute from Eq.
The above equation becomes
Solving for V and switching R to r , we then have
as the electric potential V due to a particle of charge q at any radial distance r from the particle.
A computer-generated plot of the electric potential V(r) due to a positive point charge located at the origin of a xy plane. The potentials at points in the xy plane are plotted vertically. (Curved lines have been added to help you visualize the plot.) The infinite value of V predicted by above eq. for r = 0 is not plotted.
Potential Due to a Group of Point Charges
We can find the net potential at a point due to a group of point charges with the help of the superposition principle.
Using Eq. with the sign of the charge included, we calculate separately the potential resulting from each charge at the given point. Then we sum the potentials. For n charges, the net potential is
Here qi is the value of the ith^ charge and ri is the radial distance of the given point from the ith charge. The sum in above Eq. is an algebraic sum, not a vector sum like the sum that would be used to calculate the electric field resulting from a group of point charges.
d is the distance between the charges. Under those conditions, the approximations that follow from Fig.
in which p (= qd ) is the magnitude of the electric dipole moment P. The vector P is directed along the dipole axis, from the negative to the positive charge. (Thus, θ is measured from the direction of P .) We use this vector to report the orientation of an electric dipole.
Put these value in above equation, we get
where θ is measured from the dipole axis as shown in Fig. a. We can now write V as
Potential Due to a Continuous Charge Distribution
When a charge distribution q is continuous (as on a uniformly charged thin rod or disk), we cannot use the summation to find the potential V at a point P. Instead, we must choose a differential element of charge dq , determine the potential dV at P due to dq , and then integrate over the entire charge distribution. Let us again take the zero of potential to be at infinity. If we treat the element of charge dq as a point charge, then we can express the potential dV at point P due to dq :
Here r is the distance between P and dq. To find the total potential V at P , we integrate to sum the potentials due to all the charge elements:
The integral must be taken over the entire charge distribution. Note that because the electric potential is a scalar, there are no vector components to consider. We now examine two continuous charge distributions, a line and a disk.
Treating the element as a point charge, we can write the potential dV as charge, we can write the potential dV as
Since the charge on the rod is positive and we have taken V = 0 at infinity
We now find the total potential V produced by the rod at point P by integrating above Eq. along the length of the rod, from x = 0 to x = L (Figs. d and e), using integral 17 in Appendix E. We find
We need to sum the potentials due to all the elements, from the left side (d) to the right side (e).
We can simplify this result by using the general relation ln A - ln B = ln(A/B). We then find
Because V is the sum of positive values of dV , it too is positive, consistent with the logarithm being positive for an argument greater than 1.
Charged Disk
We calculated the magnitude of the electric field at points on the central axis of a plastic disk of radius R that has a uniform charge density σ on one surface. Here we derive an expression for V(z) , the electric potential at any point on the central axis.
Note that the variable in the second integral of Eq. is Rʹ and not z, which remains constant while the integration over the surface of the disk is carried out. (Note also that, in evaluating the integral, we have assumed that z ≥ 0.)
