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A point charge of -4.00nC is at the origin, and a second point charge of 6.00nC...

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

Solutions

Expert Solution

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,

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

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

Ex = - 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,

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

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

Ex = 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,

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

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

Ex = 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.

genius_generous answered 6 months ago
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