Question

Electric field due to a line of charge The electric field in the \(x y\) -plane due to an infinite line of charge along the \(z\) -axis is a gradient field with a potential function \(V(x, y)=c \ln \left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right).\) where \(c>0\) is a constant and \(r_{0}\) is a reference distance at which the potential is assumed to be 0 (see figure). a. Find the components of the electric field in the \(x\) - and \(y\) -directions, where \(\mathbf{E}(x, y)=-\nabla V(x, y).\) b. Show that the electric field at a point in the \(x y\) -plane is directed outward from the origin and has magnitude \(|\mathbf{E}|=c / r,\) where \(r=\sqrt{x^{2}+y^{2}}.\) c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of \(V.\)

Short answer

Answer: The magnitude of the electric field at a point in the xy-plane is \(\frac{c}{r}\), where \(r = \sqrt{x^{2}+y^{2}}\).

Step-by-step solution

Find the gradient of the potential function

Given the potential function \(V(x, y) = c \ln\left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right)\), we can find the electric field by finding the negative gradient of \(V\), which is \(\mathbf{E}(x, y) = -\nabla V(x, y)\). To do this, we need to compute the partial derivatives of \(V\) with respect to \(x\) and \(y\).

Compute partial derivatives

Find the partial derivative of \(V\) with respect to \(x\): \(\frac{\partial V}{\partial x} = c \frac{\partial}{\partial x} \ln\left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right) = -\frac{c x}{x^{2}+y^{2}}\) Find the partial derivative of \(V\) with respect to \(y\): \(\frac{\partial V}{\partial y} = c \frac{\partial}{\partial y} \ln\left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right) = -\frac{c y}{x^{2}+y^{2}}\)

Find the components of the electric field

Now we can find the electric field \(\mathbf{E}(x, y)=-\nabla V(x, y)\) using the partial derivatives we found in the previous step: \(\mathbf{E}(x, y) = \left(-\frac{\partial V}{\partial x}, -\frac{\partial V}{\partial y}\right) = \left(\frac{c x}{x^{2}+y^{2}}, \frac{c y}{x^{2}+y^{2}}\right)\) Now we have the components of the electric field in the \(x\) and \(y\) directions.

Show electric field magnitude and direction

Next, we must show that the electric field at a point in the \(xy\)-plane is directed outward from the origin and has magnitude \(|\mathbf{E}| = \frac{c}{r}\), where \(r = \sqrt{x^{2}+y^{2}}\). To do this, first find \(|\mathbf{E}|\): \(|\mathbf{E}| = \sqrt{\left(\frac{c x}{x^{2}+y^{2}}\right)^{2} + \left(\frac{c y}{x^{2}+y^{2}}\right)^{2}} = \frac{c}{\sqrt{x^{2}+y^{2}}} = \frac{c}{r}\) As the electric field components are proportional to \(x\) and \(y\), they point outward from the origin.

Show the vector field is orthogonal to equipotential curves

To show that the vector field is orthogonal to the equipotential curves at all points in the domain of \(V\), we need to show that the dot product of the electric field vector and the gradient of the equipotential curve is 0. Let \(C\) be a constant, then the equipotential curve is given by: \(V(x, y) = C \implies \ln\left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right) = \frac{C}{c}\) Now, find the gradient of the equipotential curve: \(\nabla V = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}\right) = \left(-\frac{c x}{x^{2}+y^{2}}, -\frac{c y}{x^{2}+y^{2}}\right)\) Take the dot product of the electric field vector and the gradient of the equipotential curve: \(\mathbf{E} \cdot \nabla V = \left(\frac{c x}{x^{2}+y^{2}}, \frac{c y}{x^{2}+y^{2}}\right) \cdot \left(-\frac{c x}{x^{2}+y^{2}}, -\frac{c y}{x^{2}+y^{2}}\right) = 0\) The dot product is 0, which means the electric field vector and the gradient of the equipotential curve are orthogonal. We have now shown that the vector field is orthogonal to the equipotential curves at all points in the domain of \(V\).

Key concepts

Gradient Field

A gradient field is a type of vector field that represents the gradient of a scalar potential function. In physics, it is often related to how a force, such as the electric field, varies in space. For a given potential function, like the one provided in our exercise for the electric field due to a line of charge, the electric field vector can be found by taking the negative gradient of the potential function.

This requires us to understand that the gradient points in the direction of greatest increase of the function, and its magnitude gives the rate of increase per unit distance. Thus, for an electric potential function denoted by V(x, y), the gradient field, represented as abla V(x, y), will have components that correspond to the steepest ascent of potential energy in the field.

Potential Function

The potential function is a scalar function that helps calculate the electric field in electrostatics. It represents potential energy per unit charge at any point in the field. In our scenario, the potential function V(x, y) is given as a natural logarithm that depends on the position coordinates x and y.

Potential functions are very useful as they can simplify the mathematics involved in calculating forces and fields. Scalar functions are easier to work with in terms of calculus operations, such as taking derivatives, which leads us directly into computing gradient fields and understanding how they are applied in physics.

Equipotential Curves

Equipotential curves or surfaces are hypothetical lines or surfaces in a field where the potential is the same. In other words, no work is done by the field when moving a charge along these curves.

In the context of the exercise, these curves would be circles in the xy-plane centered on the z-axis, which is the line of charge. To represent these curves, we set the potential function to a constant value and solve for the spatial coordinates x and y. Equipotential curves are crucial when visualizing electric fields and understanding the nature of the forces acting at various points in space.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping the other variables constant. In the physics of electric fields, partial derivatives allow us to determine how the potential function changes in the x and y directions, which is key to finding the gradient field of that function.

As demonstrated in the exercise, by taking the partial derivatives of the potential function V with respect to x and y, we're able to calculate the electric field components corresponding to each direction. Understanding this concept allows students to tackle more complex physics problems involving fields in three-dimensional spaces.

Vector Field Orthogonality

Vector field orthogonality is a situation where two vector fields are perpendicular to each other at every point. In the context of our exercise, we are dealing with the electric field vector and the gradients of the equipotential curves. When these vectors are orthogonal, it indicates that the field direction is always perpendicular to the equipotential lines.

This orthogonality has practical implications; for instance, it means that the flow of electric force is directed in a way that does not change the potential, which is what makes equipotential surfaces so important. The mathematical condition for orthogonality is that the dot product of the two vectors is zero, a condition that the step-by-step solution we have confirms for the electric field due to a line of charge.