Answer:
r= 2.17 m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 2.9 N/C
λ = 3.5*10⁻¹⁰ C/m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2*k*(λ/r)
2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)
r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)
r= 2.17 m
The distance from the wire at which the electric field magnitude equal to 2.90 N/C is 2.17 m.
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
The intensity of the electric field due to infinite line charge is given by the following formula.
[tex]E=2k\left(\dfrac{\lambda}{r}\right)[/tex]
Here, (k) is coulombs constant,(λ) is linear charge density and (r) is the perpendicular distance of the point from the line charge.
The magnitude of electric field is 2.90 C and linear charge density is 3.50×10^−10 C/m.
As it is known that the value of coulombs constant is 8.99×10⁹ Nm²/C². Thus, the perpendicular distance of the point from the line charge is,
[tex]2.9=2(8.99\times10^{9})\left(\dfrac{3.5\times10^{-10}}{r}\right)\\r=2.17\rm\; m[/tex]
The distance from the wire at which the electric field magnitude equal to 2.90 N/C is 2.17 m.
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