Question

In: Physics

A -4.00 nCpoint charge is at the origin, and a second -6.50 nC point charge is...

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.

Solutions

Expert Solution

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]


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