Answer:
a) 7300 N/C
b) 2100 N/C
c) 5371.2 N/C
Explanation:
The uniform electric field = (4700î) N/C
The electric field due to a point charge is given as
E = kq/r²
k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²
|q| = the charge = (8.35 × 10⁻⁹) C
Note that for a negative charge, the electric field is directed from the point towards the charge.
a) x = -0.17 m.
r = (-0.17î) m
magnitude of r = 0.17 m
E = (9.0 × 10⁹ × 8.35 × 10⁻⁹)/0.17²
E = 2600 N/C
In vector form
E = (2600î) N/C (it is positive, since the field is directed from x = -0.17 m to the origin)
E(total) = 4700î + 2600î = (7300î) N/C
Magnitude = 7300 N/C
b) At x = 0.17 m
E = kq/r² gives us the same magnitude
E = 2600 N/C
But in vector form
E = (-2600î) N/C (this is because the field due to the charge is directed from x = 0.17 m to the origin, in the negative x-direction)
E(total) = 4700î - 2600î = (2100î) N/C
Magnitude = 2100 N/C
c) At y = 0.17 m
E = kq/r² is still equal in magnitude to 2600 N/C
But in vector form
E = (-2600j) N/C (directed from y = 0.17 m to the origin along the negative y-direction)
E(total) = (4700î - 2600j) N/C
Magnitude = √[4700² + (-2600)²]
Magnitude = 5371.2 N/C
Hope this Helps!!!
