Question

An electric field in free space is given as \(\mathbf{E}=x \hat{\mathbf{a}}_{x}+4 z \hat{\mathbf{a}}_{y}+4 y \hat{\mathbf{a}}_{z}\). Given \(V(1,1,1)=10 \mathrm{~V}\), determine \(V(3,3,3)\).

Short answer

Answer: The potential difference at point (3,3,3) is 82 V.

Step-by-step solution

Find \(\mathbf{E} \cdot d \boldsymbol{\ell}\)

We are given the electric field as \(\mathbf{E} = x\hat{\mathbf{a}}_{x} +4 z \hat{\mathbf{a}}_{y} + 4 y \hat{\mathbf{a}}_{z}\). Now we need to find \(d\boldsymbol{\ell}\) which is a differential length vector that lies along the path of integration. A simple choice is to take \(d\boldsymbol{\ell} = dx\hat{\mathbf{a}}_{x}+ dy\hat{\mathbf{a}}_{y}+dz\hat{\mathbf{a}}_{z}\). Thus, the dot product \(\mathbf{E} \cdot d\boldsymbol{\ell}\) can be found as $$ \mathbf{E} \cdot d\boldsymbol{\ell} = x\hat{\mathbf{a}}_{x} \cdot (dx\hat{\mathbf{a}}_{x}+ dy\hat{\mathbf{a}}_{y}+dz\hat{\mathbf{a}}_{z}) +4z\hat{\mathbf{a}}_{y}\cdot(dx\hat{\mathbf{a}}_{x}+ dy\hat{\mathbf{a}}_{y}+dz\hat{\mathbf{a}}_{z}) + 4y\hat{\mathbf{a}}_{z}\cdot (dx\hat{\mathbf{a}}_{x}+ dy\hat{\mathbf{a}}_{y}+dz\hat{\mathbf{a}}_{z}). $$ Now, we know that \(\hat{\mathbf{a}}_i \cdot \hat{\mathbf{a}}_j = \delta_{ij}\), so we get $$ \mathbf{E} \cdot d\boldsymbol{\ell} = x dx+4z dy+ 4y dz. $$

Perform the integration

We need to find the potential difference due to electric field along the path from point (1,1,1) to point (3,3,3), so we need to integrate \(\mathbf{E} \cdot d\boldsymbol{\ell}\) with respect to \(x\), \(y\), and \(z\). $$ V(3,3,3)- V(1,1,1)= -\int_{1}^{3} x dx - \int_{1}^{3} 4z dy - \int_{1}^{3} 4y dz. $$ Computing the integrals, we get $$ V(3,3,3)- V(1,1,1)= -\frac{1}{2}(3^{2} - 1^{2}) - 4 \cdot \frac{1}{2}(3^{2} - 1^{2}) - 4 \cdot \frac{1}{2}(3^{2} - 1^{2})=-8-64=-72. $$

Calculate \(V(3,3,3)\)

Now, we can find \(V(3,3,3)\) using \(V(1,1,1) = 10\) V. $$ V(3,3,3)= V(1,1,1) - (V(3,3,3)-V(1,1,1))= 10 -(-72)=82 \mathrm{~V}. $$ So, the potential difference at point (3,3,3) is \(V(3,3,3) = 82\)V.

Key concepts

Electric Field in Free Space

When studying the electric field in free space, we explore the behavior of electric forces in a vacuum or a region devoid of matter. An electric field represents the force experienced by a test charge placed within it and is a vector field characterized by both magnitude and direction.

For example, consider an electric field \textbf{E} given as \textbf{E} = x\(\bf{\hat{a}}_{x}\) + 4z\(\bf{\hat{a}}_{y}\) + 4y\(\bf{\hat{a}}_{z}\). Here, the field components relate to different coordinates, suggesting that the field's strength and direction vary depending on the position in space. To find the electric potential at a point within such a field, we apply the concept of electric potential difference which is the work done per unit charge in moving a test charge within this field.

Vector Calculus in Electromagnetism

Vector calculus is an indispensable tool in electromagnetism because fields and potentials are best described using vectors and vector operations. The dot product, for instance, is a way to quantify the work done by the electric field along a path element \(d\boldsymbol{\ell}\).

In electromagnetism, understanding how to manipulate vector fields through operations such as gradients, divergences, and curls is critical for explaining physical phenomena. For the electric field \(\mathbf{E} = x\hat{\mathbf{a}}_{x} +4 z \hat{\mathbf{a}}_{y} + 4y\hat{\mathbf{a}}_{z}\), a grasp of how these vector operations manifest in the physics is key to understanding the flow of electric charges and the potential effects they produce.

Integration of Vector Fields

Integration is a fundamental method for calculating quantities over specific regions in space or along paths. When integrating a vector field such as the electric field, we often have to deal with line integrals. In the given exercise, the line integral is used to calculate the electric potential difference between two points by integrating the component of the electric field along a chosen path.

The integration \(\int \mathbf{E} \. d\boldsymbol{\ell}\) simplifies greatly due to the orthogonality of unit vectors \(\hat{\mathbf{a}}_{i}\) (they have a dot product of zero unless they are the same), leading to separate integrals for each variable. With parameters such as x, y, and z each varying between two points, calculating the potential difference requires integrating their respective contributions, clearly shown in the steps of the provided solution.