Electric Field Calculation Uniform Line Charges Physics Problem

Introduction

In the realm of electromagnetism, understanding the electric field generated by charge distributions is crucial. This article delves into the calculation of the electric field strength produced by two uniform line charges. We will specifically address the scenario where two line charges, denoted as $[L_1]$ and $[L_2]$, each possessing a charge density of $8.854 nC/m$, are positioned along the plane $Z = 0$ at $y = +6$ m and $y = -6$ m, respectively. Our objective is to determine the electric field strength at point $P(0, 0, 6)$ due to line charge $[L_1]$. This exploration provides a foundational understanding of how to apply fundamental principles of electromagnetism to solve practical problems.

Problem Statement

Consider two infinitely long, uniform line charges, labeled as $[L_1]$ and $[L_2]$. Each line charge has a uniform charge density of $8.854 nC/m$. These line charges are situated in space along the plane defined by $Z = 0$. Specifically, $[L_1]$ is located at $y = +6$ m, and $[L_2]$ is located at $y = -6$ m. Our primary task is to determine the electric field strength, denoted as $\vec{E}$, at a specific point in space, $P(0, 0, 6)$, which is generated solely by the line charge $[L_1]$. This problem serves as a fundamental exercise in applying the principles of electrostatics to calculate the electric field due to a continuous charge distribution.

(i) Electric Field Strength Due to Line 1

To determine the electric field strength at point $P(0, 0, 6)$ due to the line charge $[L_1]$, we will employ the formula for the electric field produced by an infinitely long, uniform line charge. This formula is derived from Gauss's Law and is a cornerstone in electrostatics. The formula is given by:

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

Where:

• $\vec{E}$ represents the electric field vector.
• $\lambda$ is the linear charge density of the line charge.
• $\epsilon_0$ is the permittivity of free space, approximately $8.854 \times 10^{-12} C^2/N \cdot m^2$.
• $\rho$ is the perpendicular distance from the line charge to the point of interest.
• $\hat{\rho}$ is the unit vector pointing perpendicularly away from the line charge towards the point of interest.

In our case, $\lambda = 8.854 nC/m = 8.854 \times 10^{-9} C/m$. The line charge $[L_1]$ is located at $y = +6$ m on the Z = 0 plane, and the point of interest $P$ is at $(0, 0, 6)$. Therefore, to apply the formula effectively, we need to first calculate the perpendicular distance $\rho$ from the line charge to point $P$ and determine the direction of the unit vector $\hat{\rho}$.

The perpendicular distance $\rho$ in the x-y plane from the line charge at $(0, 6, 0)$ to the point $P(0, 0, 6)$ can be calculated using the distance formula. Since the x-coordinate of both the line charge (considering a point on the line at z=0) and point P are the same, we only need to consider the difference in the y-coordinates. The projection of point P onto the z=0 plane is (0,0,0). So, the distance in the y-direction is $|0 - 6| = 6$ m. Thus, $\rho = 6$ m. This signifies the radial distance in the x-y plane from the line charge to the point P.

Next, we need to determine the unit vector $\hat{\rho}$. This vector points from the line charge towards point $P$. The position vector from a point on the line charge $[L_1]$ (let's consider the point (0, 6, 0)) to point $P(0, 0, 6)$ is given by:

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

To find the unit vector $\hat{\rho}$, we normalize the vector $\vec{r}$ by dividing it by its magnitude:

$$|\vec{r}| = \sqrt{(-6)^2 + (6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$

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

Now, we can substitute the values of $\lambda$, $\epsilon_0$, $\rho$, and $\hat{\rho}$ into the electric field formula:

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

Simplifying the expression:

$$\vec{E} = \frac{8.854 \times 10^{-9}}{2 \pi (8.854 \times 10^{-12}) (6)} \left(-\frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z}\right) \approx 26.53 \left(-\frac{1}{\sqrt{2}}\hat{y} + \frac{1}{\sqrt{2}}\hat{z}\right)$$

$$\vec{E} \approx 26.53 \left(-0.707\hat{y} + 0.707\hat{z}\right)$$

$$\vec{E} \approx -18.76\hat{y} + 18.76\hat{z} ext{ V/m}$$

Therefore, the electric field strength at point $P(0, 0, 6)$ due to line charge $[L_1]$ is approximately $-18.76\hat{y} + 18.76\hat{z}$ V/m. This result indicates that the electric field at point $P$ has components in both the negative y-direction and the positive z-direction. The magnitude of these components is equal, suggesting that the electric field vector points at a 45-degree angle with respect to the z-axis in the y-z plane. This detailed calculation showcases the application of fundamental electrostatic principles and vector analysis in determining the electric field due to a continuous charge distribution.

Conclusion

In summary, we have successfully calculated the electric field strength at point $P(0, 0, 6)$ due to the uniform line charge $[L_1]$. By applying the formula for the electric field of an infinite line charge, derived from Gauss's Law, and carefully considering the geometry of the problem, we determined the electric field to be approximately $-18.76\hat{y} + 18.76\hat{z}$ V/m. This result underscores the importance of understanding vector components and the spatial relationships between charge distributions and points of interest. The process involved calculating the perpendicular distance from the line charge to the point, finding the unit vector pointing in the direction of the field, and then substituting these values into the electric field formula. This problem serves as a valuable example of how fundamental principles of electromagnetism can be applied to solve practical problems involving charge distributions. Further exploration might involve calculating the electric field due to the second line charge $[L_2]$ and then finding the total electric field at point $P$ by superimposing the fields from both line charges. This would provide a more complete understanding of the electric field in the region due to the combined effect of multiple charge sources.