Question

The potential function for the force field due to a charge \(q\) at the origin is \(\varphi=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{|\mathbf{r}|},\) where \(\mathbf{r}=\langle x, y, z\rangle\) is the position vector of a point in the field and \(\varepsilon_{0}\) is the permittivity of free space. a. Compute the force field \(\mathbf{F}=-\nabla \varphi\). b. Show that the field is irrotational; that is \(\nabla \times \mathbf{F}=\mathbf{0}\).

Short answer

After calculating the force field $\mathbf{F}$ and the curl of the force field, we found that the curl is equal to the zero vector: $\nabla \times \mathbf{F} = (0,0,0)$. Therefore, we can conclude that the force field is irrotational.

Step-by-step solution

Compute the gradient of the potential function

To compute the force field \(\mathbf{F}=-\nabla \varphi\), we need to calculate \(\nabla \varphi\). Recall that the gradient is given by: $$ \nabla \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right). $$ Using the given potential function \(\varphi = \frac{1}{4 \pi \varepsilon_{0}}\frac{q}{|\mathbf{r}|}\), where \(| \mathbf{r} | = \sqrt{x^2+y^2+z^2}\), we can compute the partial derivatives with respect to x, y, and z as follows: $$ \frac{\partial \varphi}{\partial x} = \frac{1}{4 \pi \varepsilon_{0}}\frac{-q x}{(x^2+y^2+z^2)^{3/2}}, \\ \frac{\partial \varphi}{\partial y} = \frac{1}{4 \pi \varepsilon_{0}}\frac{-q y}{(x^2+y^2+z^2)^{3/2}}, \\ \frac{\partial \varphi}{\partial z} = \frac{1}{4 \pi \varepsilon_{0}}\frac{-q z}{(x^2+y^2+z^2)^{3/2}}. $$

Compute the force field \(\mathbf{F}\)

Now that we have the gradient of the potential function, we can find the force field \(\mathbf{F}=-\nabla \varphi\). We simply multiply the gradient components by -1: $$ \mathbf{F} = \left( \frac{q x}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{q y}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{q z}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} \right). $$

Compute the curl of the force field \(\mathbf{F}\)

With the force field \(\mathbf{F}\), we can now compute the curl of the force field, \(\nabla \times \mathbf{F}\). The curl of a vector field is given by the determinant of the following matrix: $$ \nabla \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \\ \end{vmatrix}. $$ For \(\mathbf{F}\), this is: $$ \nabla \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{q x}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} & \frac{q y}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} & \frac{q z}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} \\ \end{vmatrix}. $$ To show that the field is irrotational, we need to show that the resulting curl is equal to \(\mathbf{0}\). When computing the curl using the determinant, we get a lengthy expression which, after simplification, results in: $$ \nabla \times \mathbf{F} = (0,0,0). $$ This shows that the force field is irrotational.

Key concepts

Force Field

A force field represents how forces are distributed across a space. Imagine a map of arrows, where each arrow shows the direction and strength of a force at different points. In our exercise, this force field is related to the potential function of a charge at the origin. The force field is derived by taking the negative gradient of the potential function. This captures how forces act around a charge, affecting any object within its vicinity.

In practice, engineers and physicists use force fields to understand interactions in electromagnetic, gravitational, and other physical systems. It's all about visualizing and quantifying how objects interact at a distance.

Gradient

The gradient is a vector that points in the direction of the steepest increase of a function. It's calculated using partial derivatives. In the context of our problem, the gradient of the potential function determines the force field.

To find the gradient, we take the derivatives of the potential function \( \varphi = \frac{1}{4 \pi \varepsilon_{0}}\frac{q}{|\mathbf{r}|} \) with respect to x, y, and z. This gives us a vector where each component reflects how the function changes in one direction:

• \( abla \varphi = \left( \frac{-q x}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{-q y}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{-q z}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} \right) \)

This provides insight into how potential changes around a charge.

Curl

The curl of a vector field measures the tendency of the field to induce rotation. It's a way to understand if and how the field curves or twists. In our exercise, calculating the curl involves using a special determinant that incorporates partial derivatives of the force field components.

For the given force field:

• \( abla \times \mathbf{F} = abla \times \left( \frac{q x}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{q y}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}}, \frac{q z}{4 \pi \varepsilon_{0}(x^2+y^2+z^2)^{3/2}} \right) \)

Simplifying the resulting expression shows the curl equals zero. This means the force field doesn't twist - it aligns with the direction of the potential gradient.

Irrotational Field

An irrotational field is a field where the curl is zero everywhere. This is a special property that indicates no rotational movement in the field. In the solution, by proving that the curl of the force field is \(\mathbf{0}\), we've shown the field is irrotational.

Irrotational fields are significant in physics because they often simplify analysis. For instance, in electrostatics and gravity, force fields around point charges are often irrotational.

• Characteristics of Irrotational Fields:
• No circulating flows occur.
• Can be described by a potential function.
• Are conservative, meaning that the work done moving around a closed loop is zero.

Recognizing an irrotational field can help in understanding the underlying physics more simply and effectively.