Respuesta :
The relationship between the electric field and the electric potential we can find the correct answers:
A) E = - (2ax + b) i^
B) E₁ = - 20.9 N / C
C) The electric field is directed towards the negative part of the x axis
D) E₂ = - 38.5 N / C
Given parameters
- The electric power V = ax² + bx + c
To find
A) Electric field
B) The magnitude of the electric field at x = 1.0 m
C) direction of electric field
D) Magnitude of the electric field at x =
The electric potential is defined as the work done by on a charge to move it within an electric field
V = - ∫ E . ds
Where bold letters indicate vectors, V is the electric potential, E the electric field and s the displacement vector.
A) To find the electric field
E = - [tex]\frac{dV}{ds}[/tex] s^
The vector s^ is a unit vector in the direction of s, in this case the electric potential is in the direction of the x axis, so the expression remains
E = - [tex]\frac{dV}{dx}[/tex] i^
Where i^ is the unit vector in the x-axis direction.
Let's do the derivative
[tex]\frac{dV}{dx}[/tex] = 2ax + b
We construct the resulting electric field vector
E = - (2ax + b) i^
We substitute the given values of the constants a and b
E = - (4.4 x + 16.5) i ^
B) They ask for the magnitude of the electric field at a point x = 1.0 m
E₁ = - (4.4 1 + 16.5)
E₁ = - 20.9 N / C
C) The electric field is directed towards the negative part of the x axis
D) The magnitude of the electric field for x = 5.0 m
E₂ = - (4.4 5 + 16.5)
E₂ = - 38.5 N / C
In conclusion, using the relationship between the electric field and the electric potential, we can find the correct answers:
A) E = - (2ax + b) i^
B) E₁ = - 20.9 N / C
C) The electric field is directed towards the negative part of the x axis
D) E₂ = - 38.5 N / C
Learn more about the electric field and electric potential here:
https://brainly.com/question/9383604
