(c) Two uniform line charges L1 and L2 of density [tex]$8.854 nC / m$[/tex] each located in space along the plane [tex]$Z =0$[/tex] at [tex]$y =+6$[/tex] and [tex]$y =-6$[/tex] m respectively. At point [tex]$P (0,0,6)$[/tex], determine the:
(i) electric field strength due to line 1 ;
(ii) electric field strength due to line 2;
(iii) total field strength.
Answer (2)
Calculate the distance r from each line to point P: r = 6 2 m .
Determine the electric field strength due to each line: E 1 = E 2 = 2 π ϵ 0 r ρ l ≈ 18.757 V/m .
Find the components of the electric field vectors E 1 and E 2 at point P: E 1 = ( − 13.263 , 0 , 13.263 ) V/m and E 2 = ( 13.263 , 0 , 13.263 ) V/m .
Calculate the total electric field strength by adding the vectors: E t o t a l = E 1 + E 2 = ( 0 , 0 , 26.526 ) V/m . The total field strength is 26.526 V/m in the z-direction.
Explanation
Problem Setup We are given two uniform line charges, L1 and L2, each with a charge density of ρ l = 8.854 nC/m . They are located at y = + 6 m and y = − 6 m respectively, both in the plane z = 0 . 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 Formula The electric field strength due to an infinite line charge is given by the formula: E = 2 π ϵ 0 r ρ l where ρ l is the line charge density, ϵ 0 is the vacuum permittivity ( 8.854 × 1 0 − 12 F/m ), and r is the distance from the line charge to the point of interest.
Calculating Distances First, we need to find the distance from each line charge to point P ( 0 , 0 , 6 ) . For line 1, located at ( 0 , 6 , 0 ) , the distance r 1 to point P is: r 1 = ( 0 − 0 ) 2 + ( 0 − 6 ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m Similarly, for line 2, located at ( 0 , − 6 , 0 ) , the distance r 2 to point P is: r 2 = ( 0 − 0 ) 2 + ( 0 − ( − 6 ) ) 2 + ( 6 − 0 ) 2 = 0 + 36 + 36 = 72 = 6 2 m Thus, r 1 = r 2 = 6 2 m .
Electric Field due to Line 1 Now we can calculate the electric field strength due to line 1 at point P: E 1 = 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 The electric field vector E 1 points from line 1 towards point P. The y-component of the position vector from line 1 to P is -6, and the z-component is 6. Thus, the angle θ between the electric field vector and the z-axis is arctan ( 6 6 ) = arctan ( 1 ) = 4 5 ∘ . The components of E 1 are: E 1 y = − E 1 sin ( 4 5 ∘ ) = − 18.757 × 2 2 ≈ − 13.263 V/m E 1 z = E 1 cos ( 4 5 ∘ ) = 18.757 × 2 2 ≈ 13.263 V/m So, E 1 = ( − 13.263 , 0 , 13.263 ) V/m .
Electric Field due to Line 2 Next, we calculate the electric field strength due to line 2 at point P: E 2 = 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 The electric field vector E 2 points from line 2 towards point P. The y-component of the position vector from line 2 to P is 6, and the z-component is 6. Thus, the angle θ between the electric field vector and the z-axis is arctan ( 6 6 ) = arctan ( 1 ) = 4 5 ∘ . The components of E 2 are: E 2 y = E 2 sin ( 4 5 ∘ ) = 18.757 × 2 2 ≈ 13.263 V/m E 2 z = E 2 cos ( 4 5 ∘ ) = 18.757 × 2 2 ≈ 13.263 V/m So, E 2 = ( 13.263 , 0 , 13.263 ) V/m .
Total Electric Field To find the total electric field strength at point P, we add the electric field vectors E 1 and E 2 :
E t o t a l = E 1 + E 2 = ( − 13.263 + 13.263 , 0 + 0 , 13.263 + 13.263 ) = ( 0 , 0 , 26.526 ) V/m The total electric field strength at point P is approximately 26.526 V/m in the z-direction.
Final Answer The electric field strength due to line 1 is approximately 18.757 V/m with components E 1 = ( − 13.263 , 0 , 13.263 ) V/m .
The electric field strength due to line 2 is approximately 18.757 V/m with components E 2 = ( 13.263 , 0 , 13.263 ) V/m .
The total electric field strength at point P is approximately 26.526 V/m in the z-direction, i.e., E t o t a l = ( 0 , 0 , 26.526 ) V/m .
Examples
Understanding electric fields created by charge distributions is crucial in designing electronic devices and ensuring their proper functioning. For example, in designing a capacitor, engineers need to calculate the electric field between the plates to optimize the device's charge storage capacity. Similarly, in telecommunications, understanding the electric fields generated by antennas is essential for efficient signal transmission and reception. This problem illustrates a fundamental concept used in these applications.
The electric field strength due to line 1 is approximately (-13.263, 0, 13.263) V/m, and due to line 2 is (13.263, 0, 13.263) V/m. The total electric field strength at point P is approximately (0, 0, 26.526) V/m. This shows the influence of both line charges on the electric field at point P.
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