Question

A point charge of -4.00nC is at the origin, and a second point charge of 6.00nC is on the x axis at x = 0.830m . Find the magnitude and direction of the electric field at each of the following points on the x axis.

a. x= 16.0 cm

b. x = 1.20m

c. x = -22.0cm

Answer 1

Charges and their positions:

q₁ = -4 * 10^-9 C     at     x₁ = 0 m

q₂ = 6 * 10^-9 C      at     x₂ = 0.83 m

k = 8.99 * 10⁹ N.m² / C² is the Coulomb's force constant

a. x = 0.16 m

X component of the electric field is,

E_x = kq₁ cos 180/ x² + kq₁ cos 180/ (0.83 - x)²

E_x = -8.99 * 10⁹ [4 * 10^-9 / (0.16)² + 6 * 10^-9 / (0.67)²]

E_x = - 1524.848 N/C

Since the charges are positioned on the x axis, there is no Y component and the magnitude of the electric field is equal to the x component sans the negative sign.

Thus,

E = 1524.848 N/C

b. This field at point x is directed in -x direction.

c. x = 1.20 m

X component of the electric field is,

E_x = kq₁ cos 180/ x² + kq₁ cos 0/ (x - 0.83)²

E_x = -8.99 * 10⁹ [4 * 10^-9 / (1.2)² - 6 * 10^-9 / (0.37)²]

E_x = 369.038 N/C

Since the charges are positioned on the x axis, there is no Y component and the magnitude of the electric field is equal to the x component.

Thus,

E = 369.038 N/C

d. This field at point x is directed in +x direction.

e. x = -0.22 m

X component of the electric field is,

E_x = kq₁ cos 0/ x² + kq₁ cos 180/ (x - 0.83)²

E_x = 8.99 * 10⁹ [4 * 10^-9 / (-0.22)² - 6 * 10^-9 / (-1.05)²]

E_x = 694.050 N/C

Since the charges are positioned on the x axis, there is no Y component and the magnitude of the electric field is equal to the x component.

Thus,

E = 694.050 N/C

f. This electric field at point x is directed in +x direction.