Understanding the electric field intensity at a specific point due to various charge distributions is a fundamental concept in electromagnetics. In this problem, we are asked to find the electric field intensity (\(E\)) at a point \(P_A (0, 0, 4)\) due to infinite uniform line charges located along the x and y axes in free space. Each line charge has a density of \(5 \text{ nC/m}\).

Electric Field from Infinite Line Charges

The electric field (\(E\)) produced by an infinite line charge with uniform charge density \(\rho_L\) at a perpendicular distance \(r\) from the line is given by the formula:

\[ E = \frac{\rho_L}{2 \pi \epsilon_0 r} \hat{a}_r \]

Where:

• \(\rho_L\) is the uniform line charge density.
• \(\epsilon_0\) is the permittivity of free space, approximately \(8.854 \times 10^{-12} \text{ F/m}\).
• \(r\) is the perpendicular distance from the line charge to the point where the electric field is being calculated.
• \(\hat{a}_r\) is the unit vector pointing radially outward from the line charge to the point.

We know that \(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\). Therefore, \(\frac{1}{2 \pi \epsilon_0} = 2 \times 9 \times 10^9 = 18 \times 10^9 \text{ Nm}^2/\text{C}^2\).

Given values:

• Line charge density, \(\rho_L = 5 \text{ nC/m} = 5 \times 10^{-9} \text{ C/m}\).
• Point where electric field is to be found, \(P_A (0, 0, 4)\).

Line Charge Distribution Analysis

The problem states that infinite uniform line charges lie along the positive and negative x and y axes. This phrasing typically refers to two distinct infinite line charges:

• One infinite line charge along the entire x-axis.
• One infinite line charge along the entire y-axis.

Both of these lines have the same charge density of \(5 \text{ nC/m}\).

Electric Field Calculation for Each Line

Electric Field from the X-axis Line Charge (\(E_x\))

The line charge is along the x-axis. The point of interest is \(P_A (0, 0, 4)\).

• The perpendicular distance \(r_x\) from the x-axis to the point \((0, 0, 4)\) is the distance from any point on the x-axis \((x, 0, 0)\) to \((0, 0, 4)\). The shortest (perpendicular) distance is from the origin \((0,0,0)\) to \((0,0,4)\) which is \(4 \text{ m}\).
• The electric field will be directed perpendicular to the x-axis and away from it. Since the point is on the positive z-axis relative to the x-axis, the direction will be \(\hat{a}_z\).

Therefore, the electric field component due to the x-axis line charge is:

\[ \vec{E}_x = \frac{\rho_L}{2 \pi \epsilon_0 r_x} \hat{a}_z \]

Substituting the values:

\[ \vec{E}_x = (5 \times 10^{-9} \text{ C/m}) \times (18 \times 10^9 \text{ Nm}^2/\text{C}^2) \times \frac{1}{4 \text{ m}} \hat{a}_z \]

\[ \vec{E}_x = \frac{90}{4} \hat{a}_z = 22.5 \hat{a}_z \text{ V/m} \]

Electric Field from the Y-axis Line Charge (\(E_y\))

The line charge is along the y-axis. The point of interest is \(P_A (0, 0, 4)\).

• The perpendicular distance \(r_y\) from the y-axis to the point \((0, 0, 4)\) is the distance from any point on the y-axis \((0, y, 0)\) to \((0, 0, 4)\). The shortest (perpendicular) distance is from the origin \((0,0,0)\) to \((0,0,4)\) which is \(4 \text{ m}\).
• Similar to the x-axis line, the electric field will be directed perpendicular to the y-axis and away from it. Since the point is on the positive z-axis relative to the y-axis, the direction will be \(\hat{a}_z\).

Therefore, the electric field component due to the y-axis line charge is:

\[ \vec{E}_y = \frac{\rho_L}{2 \pi \epsilon_0 r_y} \hat{a}_z \]

Substituting the values:

\[ \vec{E}_y = (5 \times 10^{-9} \text{ C/m}) \times (18 \times 10^9 \text{ Nm}^2/\text{C}^2) \times \frac{1}{4 \text{ m}} \hat{a}_z \]

\[ \vec{E}_y = \frac{90}{4} \hat{a}_z = 22.5 \hat{a}_z \text{ V/m} \]

Total Electric Field Intensity

According to the superposition principle, the total electric field intensity at point \(P_A\) is the vector sum of the electric fields produced by each individual line charge.

\[ \vec{E}_{total} = \vec{E}_x + \vec{E}_y \]

\[ \vec{E}_{total} = 22.5 \hat{a}_z \text{ V/m} + 22.5 \hat{a}_z \text{ V/m} \]

\[ \vec{E}_{total} = 45 \hat{a}_z \text{ V/m} \]

The total electric field intensity at \(P_A (0, 0, 4)\) is \(45 \hat{a}_z \text{ V/m}\).

This result matches one of the given options.