Calculating Electric Field Intensity From Two Uniform Line Charges

In electromagnetics, understanding the electric field intensity generated by charge distributions is crucial for various applications. This article delves into the calculation of electric field intensity at a specific point due to two uniform line charges. We will explore the concepts, formulas, and step-by-step approach to solve this problem. This is a fundamental problem in engineering, especially in fields like electrical engineering and physics, where understanding electric fields is critical for designing and analyzing electrical systems. Accurate calculation of the electric field is essential for predicting the behavior of charged particles and optimizing the performance of devices that rely on electric fields. The scenario involves two infinitely long lines of charge, each carrying a uniform charge density. These lines are positioned parallel to each other and equidistant from the z-axis. We aim to determine the net electric field intensity at a point located on the z-axis due to the combined effect of these two line charges. The problem highlights the principle of superposition, which states that the total electric field at a point is the vector sum of the electric fields due to individual charges or charge distributions. By understanding this principle, we can tackle more complex scenarios involving multiple charges or continuous charge distributions. This article will first introduce the theoretical background required to solve the problem, including the formula for the electric field due to an infinite line charge. Next, we will describe the problem setup and the location of the point at which we want to calculate the electric field intensity. Finally, we will present a step-by-step solution that applies the principle of superposition to find the net electric field intensity at the specified point. The solution will involve calculating the electric field contribution from each line charge individually and then adding them vectorially to obtain the total electric field. Throughout the article, we will emphasize the importance of understanding the underlying concepts and applying them correctly to solve problems in electromagnetics. The techniques and methods discussed here can be extended to analyze a wide range of charge configurations and calculate the electric field intensity in various situations.

Problem Statement

Consider two uniform line charges, denoted as L1 and L2, each possessing a linear charge density of $8.854 nC/m$. These line charges are situated in free space along the plane Z = 0, with their positions defined by y = +6 m and y = -6 m, respectively. Our objective is to determine the electric field intensity at point P (0, 0, 6) m, resulting from the combined effect of these two line charges. This problem is a classic example of applying the principles of electromagnetism to calculate the electric field generated by a distribution of charges. Specifically, it involves understanding the concept of linear charge density, the electric field due to an infinite line charge, and the principle of superposition. The key challenge in this problem is to correctly apply the formula for the electric field due to an infinite line charge and then perform the vector addition of the electric fields generated by the two line charges. To solve this problem, we will first calculate the electric field intensity due to each line charge individually at the given point P. This involves determining the distance from each line charge to point P and the direction of the electric field. The electric field due to an infinite line charge is radially outward (or inward, depending on the sign of the charge density) from the line charge. Therefore, the electric fields due to the two line charges at point P will have components in both the y and z directions. The next step is to resolve the electric fields due to each line charge into their y and z components. This involves using trigonometry to find the components of the electric field vectors along the respective axes. Once we have the components of the electric fields, we can add the corresponding components to find the net electric field at point P. The x-components of the electric fields will cancel out due to symmetry, leaving only the y and z components to be considered. Finally, we can calculate the magnitude and direction of the net electric field at point P using the Pythagorean theorem and the arctangent function. This will give us a complete description of the electric field intensity at the specified point due to the two uniform line charges. The solution to this problem will demonstrate the application of fundamental concepts in electromagnetism and provide a valuable understanding of how to calculate electric fields due to charge distributions.

Theoretical Background

Before diving into the solution, it's crucial to understand the underlying theoretical principles. The electric field intensity (E) due to an infinite line charge with uniform charge density (λ) at a distance (r) is given by:

$$E = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}$$

where:

• λ is the linear charge density (C/m)
• ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m)
• r is the perpendicular distance from the line charge to the point of interest
• $\hat{r}$ is the unit vector pointing radially outward from the line charge

This formula is derived from Gauss's law, a fundamental principle in electromagnetism. Gauss's law states that the electric flux through any closed surface is proportional to the enclosed electric charge. By applying Gauss's law to a cylindrical surface enclosing the line charge, we can derive the formula for the electric field intensity. The electric field intensity is a vector quantity, which means it has both magnitude and direction. The magnitude of the electric field intensity is given by the formula above, and the direction is radially outward from the line charge if the charge density is positive and radially inward if the charge density is negative. The principle of superposition is another key concept that we will use to solve this problem. The principle of superposition states that the total electric field at a point due to multiple charges is the vector sum of the electric fields due to each individual charge. This means that we can calculate the electric field due to each line charge separately and then add them together vectorially to find the total electric field. In this problem, we have two line charges, so we will calculate the electric field due to each line charge at point P and then add them together vectorially. This involves resolving the electric field vectors into their components along the x, y, and z axes and then adding the corresponding components. The resulting vector will be the total electric field at point P. Understanding these theoretical concepts is essential for solving problems in electromagnetism. The formula for the electric field due to an infinite line charge and the principle of superposition are fundamental tools that can be used to analyze a wide range of charge configurations and calculate the electric field intensity in various situations. By mastering these concepts, you will be well-equipped to tackle more complex problems in electromagnetics.

Solution

1. Electric Field Intensity due to L1

The line charge L1 is located at (0, 6, 0) and point P is at (0, 0, 6). The distance (r₁) between L1 and P is calculated as:

$$r_1 = \sqrt{(0-0)^2 + (0-6)^2 + (6-0)^2} = \sqrt{72} = 6\sqrt{2} \, m$$

The vector pointing from L1 to P is:

$$\vec{r_1} = (0-0)\hat{x} + (0-6)\hat{y} + (6-0)\hat{z} = -6\hat{y} + 6\hat{z}$$

The unit vector in this direction is:

$$\hat{r_1} = \frac{\vec{r_1}}{|\vec{r_1}|} = \frac{-6\hat{y} + 6\hat{z}}{6\sqrt{2}} = \frac{-1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z}$$

Now, we can calculate the electric field intensity (E₁) due to L1 at point P:

$$E_1 = \frac{\lambda}{2 \pi \epsilon_0 r_1} \hat{r_1} = \frac{8.854 \times 10^{-9}}{2 \pi (8.854 \times 10^{-12}) (6\sqrt{2})} \left( \frac{-1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right)$$

$$E_1 \approx 10.6 \left( \frac{-1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right) \, V/m$$

The calculation of the electric field intensity due to L1 involves several steps, each building upon the previous one. First, we need to determine the distance between the line charge L1 and the point P where we want to calculate the electric field. This distance is crucial because it appears in the denominator of the formula for the electric field due to an infinite line charge. The distance is calculated using the distance formula, which is derived from the Pythagorean theorem. Next, we need to find the vector pointing from L1 to P. This vector represents the direction of the electric field at point P due to L1. The electric field due to a positive line charge points radially outward from the line charge, so the vector pointing from L1 to P gives us the direction of the electric field. To obtain the unit vector in this direction, we divide the vector by its magnitude. The unit vector has a magnitude of 1 and points in the same direction as the original vector. Finally, we can plug the values of the linear charge density, the permittivity of free space, the distance, and the unit vector into the formula for the electric field due to an infinite line charge. This gives us the electric field intensity E₁ due to L1 at point P. The calculation involves careful attention to units and the correct application of the formula. The resulting electric field intensity is a vector quantity, with components in both the y and z directions. The negative sign in the y-component indicates that the electric field points in the negative y-direction, while the positive sign in the z-component indicates that the electric field points in the positive z-direction.

2. Electric Field Intensity due to L2

The line charge L2 is located at (0, -6, 0). The distance (r₂) between L2 and P is the same as r₁ due to symmetry:

$$r_2 = 6\sqrt{2} \, m$$

The vector pointing from L2 to P is:

$$\vec{r_2} = (0-0)\hat{x} + (0-(-6))\hat{y} + (6-0)\hat{z} = 6\hat{y} + 6\hat{z}$$

The unit vector in this direction is:

$$\hat{r_2} = \frac{\vec{r_2}}{|\vec{r_2}|} = \frac{6\hat{y} + 6\hat{z}}{6\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z}$$

Now, we calculate the electric field intensity (E₂) due to L2 at point P:

$$E_2 = \frac{\lambda}{2 \pi \epsilon_0 r_2} \hat{r_2} = \frac{8.854 \times 10^{-9}}{2 \pi (8.854 \times 10^{-12}) (6\sqrt{2})} \left( \frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right)$$

$$E_2 \approx 10.6 \left( \frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right) \, V/m$$

The calculation of the electric field intensity due to L2 follows a similar process to that for L1, taking advantage of the symmetry in the problem. Since L2 is located at (0, -6, 0) and point P is at (0, 0, 6), the distance between L2 and P is the same as the distance between L1 and P, which we calculated earlier as 6√2 meters. This is because the two line charges are equidistant from the z-axis, and point P is located on the z-axis. Next, we determine the vector pointing from L2 to P. This vector represents the direction of the electric field at point P due to L2. The electric field due to a positive line charge points radially outward from the line charge, so the vector pointing from L2 to P gives us the direction of the electric field. To obtain the unit vector in this direction, we divide the vector by its magnitude. The unit vector has a magnitude of 1 and points in the same direction as the original vector. Finally, we can plug the values of the linear charge density, the permittivity of free space, the distance, and the unit vector into the formula for the electric field due to an infinite line charge. This gives us the electric field intensity E₂ due to L2 at point P. The calculation involves careful attention to units and the correct application of the formula. The resulting electric field intensity is a vector quantity, with components in both the y and z directions. The positive sign in both the y and z components indicates that the electric field points in both the positive y-direction and the positive z-direction. This is consistent with the fact that L2 is located in the negative y-direction relative to point P, so the electric field due to L2 will have a component in the positive y-direction.

3. Total Electric Field Intensity

Using the principle of superposition, the total electric field intensity (E) at point P is the vector sum of E₁ and E₂:

$$E = E_1 + E_2 \approx 10.6 \left( \frac{-1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right) + 10.6 \left( \frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z} \right)$$

$$E \approx 10.6 \left( \frac{-1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \hat{y} + 10.6 \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \hat{z}$$

$$E \approx 0\hat{y} + 10.6 \left( \frac{2}{\sqrt{2}} \right) \hat{z}$$

$$E \approx 10.6 \sqrt{2} \hat{z} \, V/m$$

$$E \approx 15.0 \hat{z} \, V/m$$

The principle of superposition is a cornerstone of electromagnetism, allowing us to calculate the total electric field at a point due to multiple charges by simply adding the electric fields due to each individual charge. This principle simplifies the analysis of complex charge configurations and is essential for solving many problems in electromagnetics. In this case, we have calculated the electric field intensity due to each line charge, L1 and L2, at point P. Now, to find the total electric field intensity at point P, we simply add the electric field vectors E₁ and E₂. The addition of vectors involves adding their corresponding components. In this case, we add the y-components of E₁ and E₂ to get the y-component of the total electric field, and we add the z-components of E₁ and E₂ to get the z-component of the total electric field. Notice that the y-components of E₁ and E₂ have opposite signs, which means they will cancel each other out when added together. This is a result of the symmetry in the problem: the two line charges are located symmetrically about the z-axis, and point P is located on the z-axis. Therefore, the electric fields due to the two line charges will have equal and opposite y-components, resulting in a net y-component of zero. The z-components of E₁ and E₂, on the other hand, have the same sign and will add together constructively. This means that the total electric field at point P will be directed along the positive z-axis. The magnitude of the total electric field is approximately 15.0 V/m. This value represents the strength of the electric field at point P due to the combined effect of the two line charges. The direction of the electric field is along the positive z-axis, which means that a positive test charge placed at point P would experience a force in the positive z-direction.

Conclusion

The electric field intensity at point P (0, 0, 6) m due to the two uniform line charges L1 and L2 is approximately $15.0 \hat{z} , V/m$. This solution demonstrates the application of fundamental concepts in electromagnetics, including the electric field due to an infinite line charge and the principle of superposition. Understanding these concepts is crucial for solving various problems in engineering and physics related to electric fields and charge distributions. The problem we have solved is a fundamental example of how to calculate the electric field generated by a distribution of charges. By applying the formula for the electric field due to an infinite line charge and the principle of superposition, we were able to determine the net electric field at a specific point in space. This approach can be extended to analyze more complex charge configurations and calculate the electric field intensity in various situations. The solution also highlights the importance of symmetry in simplifying electromagnetic problems. In this case, the symmetry of the problem allowed us to quickly deduce that the y-components of the electric fields due to the two line charges would cancel each other out, leaving only the z-component to be considered. Recognizing and exploiting symmetry can significantly reduce the amount of calculation required to solve a problem. Furthermore, this problem provides a practical application of vector addition. The electric field is a vector quantity, and to find the total electric field due to multiple charges, we must add the electric field vectors. This involves resolving the vectors into their components along the coordinate axes and then adding the corresponding components. Vector addition is a fundamental tool in electromagnetism and is essential for solving many problems involving electric and magnetic fields. In conclusion, the calculation of the electric field intensity due to two uniform line charges is a valuable exercise in understanding and applying fundamental concepts in electromagnetics. The techniques and methods discussed here can be used to analyze a wide range of charge configurations and calculate the electric field intensity in various situations. By mastering these concepts, you will be well-equipped to tackle more complex problems in electromagnetics and electrical engineering.