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7.4: Determining Field from Potential

Recall that we were able, in certain systems, to calculate the potential by integrating over the electric field. As you may already suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need \(\vec{E}\) to calculate the force in a system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate V and then derive \(\vec{E}\) from it.

In general, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of \(\vec{E}\) and also in the direction of lower potential V. Furthermore, the magnitude of \(\vec{E}\) equals the rate of decrease of V with distance. The faster V decreases over distance, the greater the electric field. This gives us the following result.

Note: Relationship between Voltage and Uniform Electric Field

In equation form, the relationship between voltage and uniform electric field is

\[E = - \frac{\Delta V}{\Delta s}\]

where \(\Delta s\) is the distance over which the change in potential \(\Delta V\) takes place. The minus sign tells us that E points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

For continually changing potentials, \(\Delta V\) and \(\Delta s\) become infinitesimals, and we need differential calculus to determine the electric field. As shown in Figure, if we treat the distance \(\Delta s\) as very small so that the electric field is essentially constant over it, we find that

\[E_s = - \frac{dV}{ds}.\]

The figure shows the electric field component of two points A and B separated by distance delta s and having a potential difference of delta V.

Figure 7.4.1. The electric field component along the displacement \(\Delta s\) is given by \(E = - \frac{\Delta V}{\Delta s}\). Note that A and B are assumed to be so close together that the field is constant along \(\Delta s\).

Therefore, the electric field components in the Cartesian directions are given by

Note

\[E_x = - \frac{\partial V}{\partial x}, \space E_y = - \frac{\partial V}{\partial y}, \space E_z = - \frac{\partial V}{\partial z}.\]

This allows us to define the “grad” or “del” vector operator, which allows us to compute the gradient in one step. In Cartesian coordinates, it takes the form

Note

\[\vec{\nabla} = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}.\]

With this notation, we can calculate the electric field from the potential with

Note

\[\vec{E} = - \vec{\nabla}V,\]

a process we call calculating the gradient of the potential.

If we have a system with either cylindrical or spherical symmetry, we only need to use the del operator in the appropriate coordinates:

Note

\[Cylindrical: \space \vec{\nabla} = \hat{r} \frac{\partial}{\partial r} + \hat{\varphi}\frac{1}{r} \frac{\partial}{\partial \varphi} + \hat{z} \frac{\partial}{\partial z}\]

Note

\[Spherical: \space \vec{\nabla} = \hat{r} \frac{\partial}{\partial r} + \hat{\theta}\frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\varphi} \frac{1}{r \space sin \space \theta}\frac{\partial}{\partial \varphi}\]

Example

Electric Field of a Point Charge

Calculate the electric field of a point charge from the potential.

Strategy

The potential is known to be \(V = k\frac{q}{r}\), which has a spherical symmetry. Therefore, we use the spherical del operator in the formula \(\vec{E} = - \vec{\nabla}V\).

Solution

Performing this calculation gives us

\[\vec{E} = - \left( \hat{r}\frac{\partial}{\partial r} + \hat{\theta}\frac{1}{r} \frac{\partial}{\partial \theta} + \hat{\varphi}\frac{1}{1 \space sin \space \theta} \frac{\partial}{\partial \varphi}\right) k\frac{q}{r} = - k\left( \hat{r}\frac{\partial}{\partial r}\frac{1}{r} + \hat{\theta}\frac{1}{r} \frac{\partial}{\partial \theta}\frac{1}{r} + \hat{\varphi}\frac{1}{1 \space sin \space \theta} \frac{\partial}{\partial \varphi}\frac{1}{r}\right).\]

This equation simplifies to

\[\vec{E} = - kq\left(\hat{r}\frac{-1}{r^2} + \hat{\theta}0 = \hat{\varphi}0 \right) = k\frac{q}{r^2}\hat{r}\] as expected.

Significance

We not only obtained the equation for the electric field of a point particle that we’ve seen before, we also have a demonstration that \(\vec{E}\) points in the direction of decreasing potential, as shown in Figure.

The figure shows a charge Q and radially outward electric field vectors from Q.

Figure 7.4.2. Electric field vectors inside and outside a uniformly charged sphere.

Example

Electric Field of a Ring of Charge Use the potential found in [link] to calculate the electric field along the axis of a ring of charge (Figure).

The figure shows a ring of charge located on the xy-plane with its center at the origin. Point P is located on the z-axis at distance z away from the origin.

Figure 7.4.3. We want to calculate the electric field from the electric potential due to a ring charge.

Strategy

In this case, we are only interested in one dimension, the z-axis. Therefore, we use \(E_z = - \frac{\partial V}{\partial z}\) with the potential \(V = k \frac{q_{tot}}{\sqrt{z^2 + R^2}}\) found previously.

Solution

Taking the derivative of the potential yields

\[E_z = - \frac{\partial}{\partial z} \frac{kq_{tot}}{\sqrt{z^2 + R^2}} = k \frac{q_{tot}z}{(z^2 + R^2)^{3/2}}.\]

Significance

Again, this matches the equation for the electric field found previously. It also demonstrates a system in which using the full del operator is not necessary.

Note

Check Your Understanding Which coordinate system would you use to calculate the electric field of a dipole?

[Hide Solution]

Any, but cylindrical is closest to the symmetry of a dipole.

Summary

  • Just as we may integrate over the electric field to calculate the potential, we may take the derivative of the potential to calculate the electric field.
  • This may be done for individual components of the electric field, or we may calculate the entire electric field vector with the gradient operator.

Conceptual Questions

Exercise

If the electric field is zero throughout a region, must the electric potential also be zero in that region?

[Hide Solution]

No. It will be constant, but not necessarily zero.

Exercise

Explain why knowledge of \(\vec{E}(x,y,z)\) is not sufficient to determine V(x,y,z). What about the other way around?

Problems

Exercise

Throughout a region, equipotential surfaces are given by \(z = constant\). The surfaces are equally spaced with \(V = 100 \space V\) for \(z = 0.00 \space m\), \(V = 200 \space V\) for \(z = 0.50 \space m\), \(V = 300 \space V\) for \(z = 1.00 \space m\). What is the electric field in this region?

[Hide Solution]

The problem is describing a uniform field, so \(E = 200 \space V/m\) in the –z-direction.

Exercise

In a particular region, the electric potential is given by \(V = - xy^2z + 4xy\). What is the electric field in this region?

Exercise

Calculate the electric field of an infinite line charge, throughout space.

[Hide Solution]

Apply \(\vec{E} = - \vec{\nabla}V\) with \(\vec{\nabla} = \hat{r}\frac{\partial}{\partial r} + \hat{\varphi} \frac{1}{r} \frac{\partial}{\partial \varphi} + \hat{z}\frac{\partial}{\partial z}\) to the potential calculated earlier,

\(V = -2k\lambda \space ln \space s: \space \vec{E} = 2k\frac{1}{r}\hat{r}\) as expected.