(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 electric field E 1 due to line 1 at point P, finding the distance r 1 = 6 2 m and the unit vector r 1 ^ = ( 0 , − 2 1 , 2 1 ) , resulting in E 1 ≈ ( 0 , − 13.26 , 13.26 ) V/m .
Calculate the electric field E 2 due to line 2 at point P, finding the distance r 2 = 6 2 m and the unit vector r 2 ^ = ( 0 , 2 1 , 2 1 ) , resulting in E 2 ≈ ( 0 , 13.26 , 13.26 ) V/m .
Add the electric fields E 1 and E 2 to find the total electric field E t o t a l = E 1 + E 2 .
The total electric field strength at point P is ( 0 , 0 , 26.52 ) V/m .
Explanation
Problem Setup We are given two uniform line charges, L 1 and L 2 , each with a charge density of ρ l = 8.854 nC/m . They are located at y = + 6 m and y = − 6 m respectively, along the z = 0 plane. 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 - Distance and Unit Vector First, let's calculate the electric field strength due to line 1 ( E 1 ). The distance from line 1 to point P is r 1 = ( 0 − 0 ) 2 + ( 0 − 6 ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m . The unit vector pointing from line 1 to point P is r 1 ^ = 6 2 ( 0 , − 6 , 6 ) = ( 0 , − 2 1 , 2 1 ) .
Calculating Electric Field Strength E1 The electric field strength due to line 1 is given by E 1 = 2 π ϵ 0 r 1 ρ l r 1 ^ , where ϵ 0 = 8.854 × 1 0 − 12 F/m . Plugging in the values, we have
E 1 = 2 π ( 8.854 × 1 0 − 12 ) ( 6 2 ) 8.854 × 1 0 − 9 ( 0 , − 2 1 , 2 1 ) = 2 π ( 6 2 ) × 1 0 − 3 1 ( 0 , − 2 1 , 2 1 )
We found that the magnitude 2 π ϵ 0 r 1 ρ l = 18.75658991993971 . Therefore,
E 1 = 18.7566 ( 0 , − 2 1 , 2 1 ) ≈ ( 0 , − 13.26 , 13.26 ) V/m
Electric Field due to Line 2 - Distance and Unit Vector Next, let's calculate the electric field strength due to line 2 ( E 2 ). The distance from line 2 to point P is r 2 = ( 0 − 0 ) 2 + ( 0 − ( − 6 ) ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m . The unit vector pointing from line 2 to point P is r 2 ^ = 6 2 ( 0 , 6 , 6 ) = ( 0 , 2 1 , 2 1 ) .
Calculating Electric Field Strength E2 The electric field strength due to line 2 is given by E 2 = 2 π ϵ 0 r 2 ρ l r 2 ^ . Plugging in the values, we have
E 2 = 2 π ( 8.854 × 1 0 − 12 ) ( 6 2 ) 8.854 × 1 0 − 9 ( 0 , 2 1 , 2 1 ) = 2 π ( 6 2 ) × 1 0 − 3 1 ( 0 , 2 1 , 2 1 )
We found that the magnitude 2 π ϵ 0 r 2 ρ l = 18.75658991993971 . Therefore,
E 2 = 18.7566 ( 0 , 2 1 , 2 1 ) ≈ ( 0 , 13.26 , 13.26 ) V/m
Total Electric Field Strength Finally, the total electric field strength at point P is the sum of the electric fields due to line 1 and line 2: E t o t a l = E 1 + E 2 = ( 0 , − 13.26 , 13.26 ) + ( 0 , 13.26 , 13.26 ) = ( 0 , 0 , 26.52 ) V/m .
Final Answer Therefore, the electric field strength due to line 1 is approximately ( 0 , − 13.26 , 13.26 ) V/m , the electric field strength due to line 2 is approximately ( 0 , 13.26 , 13.26 ) V/m , and the total electric field strength at point P is approximately ( 0 , 0 , 26.52 ) V/m .
Examples
Understanding electric fields from 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 power lines as line charges, they can predict the electric field at various distances and design the system to meet safety standards. This ensures that the electric fields do not exceed the air's dielectric strength, preventing dangerous arcing or corona discharge.
