Respuesta :
The magnitude and direction of the electric field at a distance of 1.50 m from a 50.0-NC charge is option(d) i.e,200 N/C away from the charge.
An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field.
Equation E = kQ/[tex]r^{2}[/tex] determines the size of the electric field (E) created by a point charge with a charge of magnitude Q at a position that is r distant from the point charge. k is a constant with a value of 8.99 x [tex]10^{9}[/tex] N [tex]m^{2}[/tex][tex]C^{2}[/tex].
E = kQ/[tex]r^{2}[/tex]
E = 8.99 x[tex]10^{9}[/tex] × 50 × [tex]10^{-9}[/tex]/[tex]1.50^{2}[/tex]
E = 199.8N/C = 200N/C
As the charge is positive the direction of the electric field will be away from the charge. Therefore, the correct answer is option(d)i.e,200 N/C away from the charge.
The complete question is:
What are the magnitude and direction of the electric field at a distance of 1.50 m from a 50.0-NC charge?
A) 20 N/C away from the charge
B) 20 N/C toward from the charge
C) 10 N/C away from the charge
D) 200 N/C away from the charge
E) 200 N/C toward from the charge
To know more about electric field refer to: https://brainly.com/question/8971780
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