(c) Two uniform line charges $L_1$ and $L_2$ of density $8.854 nC/m$ each located in space along the plane $Z = 0$ at $y = +6$ and $y = -6$ m respectively. At point $P (0,0,6)$, determine the:
(i) electric field strength due to line 1 ;
(ii) electric field strength due to line 2 ;
(iii) total field strength.
Answer (1)
Calculate the distance r 1 and r 2 from each line to point P: r 1 = r 2 = 6 2 m .
Determine the unit vectors a r 1 and a r 2 from each line to point P: a r 1 = 2 − a y + a z and a r 2 = 2 a y + a z .
Calculate the electric field strength due to each line: E 1 ≈ − 13.26 a y + 13.26 a z V/m and E 2 ≈ 13.26 a y + 13.26 a z V/m .
Find the total electric field strength by adding the individual fields: E t o t a l = E 1 + E 2 ≈ 26.52 a z V/m .
Explanation
Problem Setup We are given two uniform line charges, L 1 and L 2 , each with a density of ρ l = 8.854 nC/m . They are located along the z=0 plane at y = + 6 m and y = − 6 m respectively. We want to find the electric field strength due to each line charge and the total electric field strength at point P ( 0 , 0 , 6 ) .
Electric Field due to Line 1 First, let's find the electric field strength due to line 1. The distance r 1 from line 1 to point P is given by: r 1 = ( 0 − 0 ) 2 + ( 0 − 6 ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m The unit vector a r 1 from line 1 to point P is given by: r 1 = ( 0 − 0 ) a x + ( 0 − 6 ) a y + ( 6 − 0 ) a z = − 6 a y + 6 a z a r 1 = 6 2 − 6 a y + 6 a z = 2 − a y + a z The electric field strength due to line 1 is given by: E 1 = 2 π ϵ 0 r 1 ρ l a r 1 where ϵ 0 = 8.854 × 1 0 − 12 F/m .
Electric Field due to Line 2 Now, let's find the electric field strength due to line 2. The distance r 2 from line 2 to point P is given by: r 2 = ( 0 − 0 ) 2 + ( 0 − ( − 6 ) ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m The unit vector a r 2 from line 2 to point P is given by: r 2 = ( 0 − 0 ) a x + ( 0 − ( − 6 )) a y + ( 6 − 0 ) a z = 6 a y + 6 a z a r 2 = 6 2 6 a y + 6 a z = 2 a y + a z The electric field strength due to line 2 is given by: E 2 = 2 π ϵ 0 r 2 ρ l a r 2
Calculating Electric Field Magnitudes We have ρ l = 8.854 × 1 0 − 9 C/m , ϵ 0 = 8.854 × 1 0 − 12 F/m , and r 1 = r 2 = 6 2 m . Therefore, 2 π ϵ 0 r 1 ρ l = 2 π ( 8.854 × 1 0 − 12 ) ( 6 2 ) 8.854 × 1 0 − 9 ≈ 18.757 V/m 2 π ϵ 0 r 2 ρ l = 2 π ( 8.854 × 1 0 − 12 ) ( 6 2 ) 8.854 × 1 0 − 9 ≈ 18.757 V/m So, E 1 = 18.757 2 − a y + a z ≈ − 13.26 a y + 13.26 a z V/m E 2 = 18.757 2 a y + a z ≈ 13.26 a y + 13.26 a z V/m
Total Electric Field The total electric field strength is the sum of the electric fields due to line 1 and line 2: E t o t a l = E 1 + E 2 = ( − 13.26 a y + 13.26 a z ) + ( 13.26 a y + 13.26 a z ) = 26.52 a z V/m
Final Answer Therefore, the electric field strength due to line 1 is approximately − 13.26 a y + 13.26 a z V/m , the electric field strength due to line 2 is approximately 13.26 a y + 13.26 a z V/m , and the total electric field strength at point P is approximately 26.52 a z V/m .
Examples
Understanding electric fields created by line charges is crucial in designing high-voltage power lines. Engineers need to calculate the electric field strength around these lines to ensure safety and prevent electrical breakdown in the surrounding air. By modeling the power lines as line charges, they can predict the electric field distribution and optimize the line placement and insulation to minimize potential hazards and ensure reliable power transmission.
