Question

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(V=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla V,\) where \(\nabla V\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

Short answer

Question: Calculate the magnitude of the electric field vector for the given potential function, and explain why it is referred to as an inverse square law. Solution: 1. Calculate the gradient of the potential function V = kQ/r in three dimensions. 2. Negate the gradient to find the electric field vector. 3. Determine the magnitude of the electric field vector. 4. Observe that the magnitude is inversely proportional to the square of the distance r (|E| = kQ/r^2). The term "inverse square law" refers to the fact that the magnitude of the electric field decreases with the square of the distance between the point charge and the point of interest, such that any increase in distance results in rapid decrease in electric field strength.

Step-by-step solution

Calculate the gradient of V

Firstly, we need to find the gradient of the potential function \(V\). The gradient of a function in three dimensions is given by: $$ \nabla V=\left\langle\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z}\right\rangle $$ Recall that the potential function \(V=k Q / r\), and \(r^{2}=x^{2}+y^{2}+z^{2}\). So, we will find the partial derivatives of \(V\) with respect to \(x\), \(y\), and \(z\). $$ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$ $$ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$ $$ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}\left(kQ\frac{1}{\sqrt{x^2+y^2+z^2}}\right) $$

Evaluate the partial derivatives

Now we'll evaluate the three partial derivatives. Using the chain rule, we have: $$ \frac{\partial V}{\partial x} = -kQ\frac{x}{(x^2+y^2+z^2)^{3/2}} $$ $$ \frac{\partial V}{\partial y} = -kQ\frac{y}{(x^2+y^2+z^2)^{3/2}} $$ $$ \frac{\partial V}{\partial z} = -kQ\frac{z}{(x^2+y^2+z^2)^{3/2}} $$

Find the electric field vector

Since \(\mathbf{E}=-\nabla V\), we can now find the electric field vector by plugging in the gradient components: $$ \mathbf{E}(x, y, z) = kQ \left\langle \frac{x}{r^3},\frac{y}{r^3},\frac{z}{r^3} \right\rangle $$ b. Finding the magnitude of the electric field and explaining the inverse square law.

Calculate the magnitude of the electric field vector

To find the magnitude of the electric field vector, recall the formula for the magnitude of a vector: $$ |\mathbf{E}| = \sqrt{(E_x)^2+(E_y)^2+(E_z)^2} $$ Plugging in the components of \(\mathbf{E}\), we get: $$ |\mathbf{E}| = \sqrt{\left(kQ\frac{x}{r^3}\right)^2+\left(kQ\frac{y}{r^3}\right)^2+\left(kQ\frac{z}{r^3}\right)^2} $$ Simplifying, we obtain: $$ |\mathbf{E}| = \frac{kQ}{r^2} $$

Explain why it is called an inverse square law

The relationship \(|\mathbf{E}| = kQ/r^2\) is called an inverse square law because the magnitude of the electric field \(|\mathbf{E}|\) is inversely proportional to the square of the distance \(r^2\) between the point charge and the point of interest. This means that as the distance away from the point charge doubles, the electric field's strength decreases by a factor of 4, and as the distance increases, the electric field strength decreases even more rapidly.

Key concepts

Gradient

The concept of a gradient is critically important in understanding how potential functions relate to physical fields, such as electric fields. A gradient in mathematics refers to a vector that points in the direction of the greatest rate of increase of a scalar function. In three dimensions, it is represented as:

• \( abla V = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right\rangle \)

This notation indicates that the gradient is comprised of the partial derivatives of the function with respect to each spatial coordinate. In the context of potential functions, the gradient illustrates how steeply the potential increases or decreases at a given point.
For example, the electric field is derived from the potential function as the negative of its gradient. This means that the electric field vector points in the direction that steeply decreases potential, and the magnitude of this vector indicates the strength of the field. The relationship between the potential function and the electric field vector underscores the power of gradients in translating mathematical descriptions into physical realities.

Electric Field

An electric field is a vector field around a charged object where electric forces occur. It signifies the strength and direction that a charge would move if placed in the field. The electric field \( \mathbf{E} \) due to a point charge is computed as the negative gradient of the potential function \( V \). Typically, the potential, \( V \), for a point charge is expressed as:

• \( V = \frac{kQ}{r} \)

Where \( Q \) is the charge, \( r \) is the distance from the charge, and \( k \) is a constant.
To find the electric field, calculate \( -abla V \). This results in the following expression for the electric field vector:

• \( \mathbf{E} = kQ \left\langle \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \right\rangle \)

The components \( \frac{x}{r^3}, \frac{y}{r^3}, \frac{z}{r^3} \) reveal how the electric field varies in each spatial direction. This field pattern shows that electric fields emanate outward in symmetrical patterns when looking at a charge in a vacuum. This formulation suggests that the field's direction depends on the spatial coordinates \( (x, y, z) \). Hence, they directly influence the resulting electric field through their relationship to \( r \), the radial distance.

Inverse Square Law

The inverse square law describes how the magnitude of certain physical quantities diminishes with distance. In the case of the electric field from a point charge, this law tells us how the electric force decreases as we move away from the charge. The magnitude of the electric field \( |\mathbf{E}| \) is given as:

• \( |\mathbf{E}| = \frac{kQ}{r^2} \)

This equation illustrates the essence of the inverse square law: as the distance \( r \) from the charge increases, the magnitude of \( |\mathbf{E}| \) rapidly decreases, specifically by a factor of \( 1/r^2 \).
This relationship emerges from geometric considerations. When the distance doubles, the area through which the electric field lines spread increases by a factor of four (\( 2^2 \)), causing the electric field's strength to drop to one-quarter.
Understanding the inverse square law is fundamental in physics, not just for electrical phenomena but also for gravitational and illumination applications. Such recognition of broad occurrences helps students grasp how physical laws apply in diverse contexts.