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!!!
Answer: 7700N/C, 5521.77N/C
Explanation:
Complete question
(this electric field has a magnitude of 4500 N/C and is directed in the positive x direction. A point charge -8.0 * 10^-9 C is placed at the origin. Determine the magnitude of the net electric field at
a. x= -0.15m and b. y=0.15m)
Okay negative (-ve) side the field is pointing in the + x direction which means the low potential is to the left hand side.
q = -8.0*10-⁹C
E = 4500 N/C
E = Kq/r²
Where q = 9.0E9 Nm²/c
x = r and y = r
Ec(x) = kq/x² = (9.0E9 * 8.0E-9) / (-0.15)² = 3200 N/C
1. E(total) = E(uniform) + Ec(point)
E(total) = 4500 + 3200 = 7700N/C
2. E(total) = [E(uniform)² + Ey(point)²]^½
E(total = [ 4500² + 3200²]^½
E(total) = 5521.77N
