Answer:
[tex]\mathbf{E = 3.04 \times 10^4 \ N/C}[/tex] and the direction is vertically upward
Explanation:
To find the magnitude & direction which was being produced at a given point of 6.00 cm which is directly above the midpoint.
Given that:
the wire length = 8.50 cm = 8.5 × 10⁻² m
the charge density λ = 175 n C/m = 175 ×10⁻⁹ C/m
distance x = 6.00 cm = 0.006 m
Assume that N be the point at distance x situated directly above the midpoint
Then, the linear expression for the linear charge density is:
[tex]\lambda = \dfrac{Q}{L}[/tex]
the charge [tex]Q = \lambda L[/tex]
[tex]Q = 175 \times 10^{-9} \times 8. 5 \times 10^{-2}[/tex]
[tex]Q = 1.49 \times 10^{-8} \ C[/tex]
For the electric field at point N;
[tex]E = \dfrac {kQ}{x\sqrt{x^2 + r^2}}[/tex]
where;
[tex]r = \dfrac{L} {2}[/tex]
[tex]E = \dfrac {9\times 10^9 \times 1.49 \time 10^{-8} }{0.06 \times \sqrt{0.06^2 + (\dfrac{8.50 \times 10^{-2}}{2})^2}}[/tex]
[tex]\mathbf{E = 3.04 \times 10^4 \ N/C}[/tex]