Question

A uniform line charge of \(16 \mathrm{nC} / \mathrm{m}\) is located along the line defined by \(y=\) \(-2, z=5\). If \(\epsilon=\epsilon_{0}:\) (a) find \(\mathbf{E}\) at \(P(1,2,3) .\) (b) find \(\mathbf{E}\) at that point in the \(z=0\) plane where the direction of \(\mathbf{E}\) is given by \((1 / 3) \mathbf{a}_{y}-(2 / 3) \mathbf{a}_{z} .\)

Short answer

Short Answer: (a) The electric field at point \(P(1, 2, 3)\) is: $$\mathbf{E}(1, 2, 3) = 8\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_y + 4\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_z$$ (b) The electric field at the given point on the \(z=0\) plane with the given direction is: $$\mathbf{E}(x, 10, 0) = \frac{80}{169}\,\mathrm{N/C} \, \mathbf{a}_y + \frac{40}{169}\,\mathrm{N/C} \, \mathbf{a}_z$$

Step-by-step solution

Calculate the electric field due to a line charge

To calculate the electric field due to a line charge at a given point, we use the formula: $$\mathbf{E} = \int_{l} \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_R dl$$ Where \(\lambda\) is the linear charge density, \(R\) is the distance from the line charge to the point of interest, and \(\mathbf{a}_R\) is the unit vector in the radial direction.

Determine the coordinates of the line charge at point P

We are given point \(P(1, 2, 3)\). The line charge is at \(y=-2\) and \(z=5\). We subtract the coordinates to find the difference in position. $$\Delta y = 2 - (-2) = 4$$ $$\Delta z = 5 - 3 = 2$$ Now we can find the distance and the electric field: $$ R = \sqrt{\Delta y^2 + \Delta z^2} = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5} $$ Now we can find the components of the electric field, \(\mathbf{E}_y\) and \(\mathbf{E}_z\).

Calculate the Electric Field components

The electric field components can be calculated as follows: $$ \mathbf{E}_y = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_y $$ $$ \mathbf{E}_z = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_z $$ Now plug in the values we have: $$ \mathbf{E}_y = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(2\sqrt{5})^2} \mathbf{a}_y = 8\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_y $$ $$ \mathbf{E}_z = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(2\sqrt{5})^2} \mathbf{a}_z = 4\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_z $$ So, the electric field at point P is: $$\mathbf{E}(1, 2, 3) = 8\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_y + 4\sqrt{5} \,\mathrm{N/C} \, \mathbf{a}_z$$

Calculate the Electric Field at the given point in the z=0 plane

We are given the direction of the electric field at a point on the \(z=0\) plane: $$\mathbf{E} = E_y\frac{\mathbf{a}_y}{3} - E_z\frac{2\mathbf{a}_z}{3}$$ The point is on the \(z=0\) plane, so we can write its coordinates as \((x, y, 0)\). The line charge is at \(y=-2\) and \(z=5\). We subtract the coordinates to find the difference in position. $$\Delta y = y - (-2)$$ $$\Delta z = 5 - 0 = 5$$ Now we can find the distance and the electric field: $$ R = \sqrt{\Delta y^2 + \Delta z^2} = \sqrt{(\Delta y)^2 + 5^2} $$ Now we can use the relationship between the electric field components: $$\frac{E_y}{3} = \frac{2E_z}{3}$$ Using the expression for the electric field due to a line charge, we can write: $$\frac{1}{4\pi\epsilon_0} \frac{\lambda}{(\Delta y^2 + 25)^{3/2}}\Delta y = 2 \times \frac{1}{4\pi\epsilon_0} \frac{\lambda}{(\Delta y^2 + 25)^{3/2}} 5$$ Simplifying the equation, we get: $$\Delta y = 10$$ So, the point at which the given electric field direction occurs is \((x,10,0)\). Now we can find the electric field components at this point: $$\mathbf{E}_y = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(5^2 + 10^2)^{3/2}} 10$$ $$\mathbf{E}_z = \frac{9 \times 10^9 \,\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \cdot 16 \times 10^{-9} \,\mathrm{C}/\mathrm{m}}{(5^2 + 10^2)^{3/2}} 5$$ So, the electric field at the given point on the \(z=0\) plane is: $$\mathbf{E}(x, 10, 0) = \frac{80}{169}\,\mathrm{N/C} \, \mathbf{a}_y + \frac{40}{169}\,\mathrm{N/C} \, \mathbf{a}_z$$

Key concepts

Line Charge

A line charge refers to a distribution of electric charge along a line, where the charge is spread out uniformly over the length of the line. The linear charge density, denoted by \( \lambda \), usually defined as charge per unit length \( (\mathrm{C/m}) \), is a key characteristic of a line charge. In the given exercise, a line charge of \( 16 \, \mathrm{nC/m} \) is located along the line described by two plane coordinates \( y = -2 \) and \( z = 5 \). This means that along this line in three-dimensional space, the charge is spread uniformly at a rate of \( 16 \, \mathrm{nC/m} \). Understanding this concept is crucial as line charges are a common source of electrical fields in theoretical and practical applications.

Electric Field Calculation

The electric field \( \mathbf{E} \) represents the force that a charge experiences in the presence of another electric charge. To calculate the electric field due to a line charge, we use an integral that accounts for contributions from each differential element \( dl \) of the line. The expression is given by:

• \( \mathbf{E} = \int_{l} \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2}\mathbf{a}_R \, dl \)

Here, \( \epsilon_0 \) is the permittivity of free space, \( \lambda \) is the linear charge density, \( R \) is the perpendicular distance from the line charge to the point of interest, and \( \mathbf{a}_R \) is the unit vector in the direction from the charge to this point. The key part of this calculation involves determining \( R \) and \( \mathbf{a}_R \) at the point where the field is calculated, ensuring the effects of all differential charge elements are summed properly.

Electric Field Components

Once the overall electric field is calculated, it is often essential to consider its components along specific axes, usually dictated by the geometry of the problem. In the solution provided:

• \( \mathbf{E}_y \) and \( \mathbf{E}_z \) represent the y and z components of the electric field.

For each component, the electric field due to the line charge can be computed using:

• \( \mathbf{E}_y = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2} \mathbf{a}_y \)
• \( \mathbf{E}_z = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R^2} \mathbf{a}_z \)

These formulas help break down the total electric field into parts that can be analyzed along the coordinates of interest. For example, in the problem, the value of the electric field at point \( P(1,2,3) \) was determined in terms of these components.

Coulomb's Law

Coulomb's law is fundamental in understanding interactions between charges, providing the basis for calculating electric fields due to charge distributions. It states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The force is directed along the line joining the charges. In mathematical terms, for point charges:

• \( F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} \)

where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between them, and \( \epsilon_0 \) is the permittivity of free space. Although the exercise involves a line charge, the principle behind the electric field calculation stems from an extension of this law. By integrating the effects of an infinite series of point charges along a line, we apply Coulomb's Law in a continuous manner.