Answer: 71.93 *10^3 N/C
Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:
∫E*dr=Q inside/εo Q inside is given by: λ*L then,
E*2*π*r*L=λ*L/εo
E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C
The magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 is mathematically given as
E= 71.93 * 10^3 N/C towards the positive axis
Question Parameter(s):
The magnitude and direction of the electric field 2.0 m
Charged uniformly at λ = 4.0 × 10-6 C/m.
Generally, the equation for the Gaussian law is mathematically given as
∫E*dr=Q
Therefore
E*2*π*r*L=λ*L/εo
E= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )
E= 71.93 * 10^3 N/C
In conclusion, the magnitude electric field
E= 71.93 * 10^3 N/C
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