In: Physics
A -4.00 nCpoint charge is at the origin, and a second -6.50 nC point charge is on the x-axis at x = 0.800 mm.
Part A
Find the electric field (magnitude and direction) at point on the x-axis at x = 0.200 mm.
Express your answer with the appropriate units. Enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.
Part B
Find the electric field (magnitude and direction) at point on the x-axis at x = 1.20 mm.
Express your answer with the appropriate units. Enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.
Part C
Find the electric field (magnitude and direction) at point on the x-axis at x = -0.200 mm.
Express your answer with the appropriate units. Enter positive value if the field is in the positive x-direction and negative value if the field is in the negative x-direction.
q1 =
q2 =
r1 = distance of point from q1
r2 = distance of point from q2
E1 = electric field due to q1 =
E2 = electric field due to q2. =
A)
r1 = distance of point A from q1 = 0.2 mm = 0.0002 m.
r2 = distance of point A from q2 = 0.6 mm = 0.0006 m.
E1 = .
E2 =
Net electric field = E1 - E2 = 7.38 x 108 N/C , and it's direction is towards -ve x-axis. [answer]
B)
r1 = distance of point A from q1 = 1.2 mm = 0.0012 m.
r2 = distance of point A from q2 = 0.4 mm = 0.0004 m.
E1 = .
E2 =
Net electric field = E1 + E2 = 3.85x 108 N/C , and it's direction is towards -ve x-axis and it's direction is towards -ve x-axis [answer]
C)
r1 = distance of point A from q1 = 1.2 mm = 0.0002 m.
r2 = distance of point A from q2 = 0.4 mm = 0.001 m.
E1 = .
E2 =
Net electric field = E1 + E2 = 9.585x 108 N/C , and it's direction is towards +ve x-axis and it's direction is towards +ve x-axis [answer]