A -3.00 nC point charge is at the origin, and a second -6.50 nC point charge is

Question

A -3.00 nC point charge is at the origin, and a second -6.50 nC point charge is on the z-axis at x -0.800 m. Part A Find the electric field (magnitude and direction) at point on the z-axis at x = 0.200 m. Express your answer with the appropriate units. Enter positive value if the field is in the positive -direction and negative value if the field is in the negative r-direction. B+1 Value Units SubmitRequest Answer Part B Find the electric field (magnitude and direction) at point on the z-axis at x = 1.20 m. Express your answer with the appropriate units. Enter positive value if the field is in the positive r-direction and negative value if the field is in the negative r-direction. Ez= Value Units

Explanation / Answer

superposition theorem reveal that the potential at any point in space, due to static distribution of charge, is algebraic sum of the individual potentials produced by each, individual charge.,

i.e., E(net) = E1+E2+E3 +....

E-Field produced by a charge is given by

E =Kq /r2

Here, K = 1/ (4o) = 9.0 x109

Now q1 = -3.0nC or -3.0x10-9 C

                q2 = -6.0nC or -6.5x10-9C

Let the distance from q1 to required point is noted as r1

And

The distance from q2 to required point is noted as r2.

r1 = 0.2m

r2 =0.8-0.2 =0.6m

Electric field, E(q1) = [9 x 109 x (-3 x 10-9) ] / (0.2)2

                                = -675J

Electric field, E(q2) = [9 x 109 x (-6.5 x 10-9) ] / (0.6)2

                                = -162.5J

E(net) = E(q1) + E(q2) =-675J + -162.5J = -837.5J [the negative sign means it points in the negative x- direction]

b)

r1 = 1.2m

r2 = (1.2 – 0.8) = 0.4m

E(q1) = [9 x 109 x (-3 x 10-9) ] / (1.2)2

= -18.75J

E(q2) = [9 x 109 x (-6.5 x 10-9) ] / (0.4)2
            = -365.625J

E(net) = E(q1) + E(q2) =-18.75J + -365.625 = -384.375J [(-) sign means it points in the negative x- direction]

c)

r1 = (0.2+0.2) = 0.4m
r2 = (0.8+0.2) = 1.0m

E(q1) = [9 x 109 x (-3.0 x 10-9) ] / (0.4)2

                = -168.75J

E(q2) = [9 x 109 x (-6.5 x 10-9) ] / (1.0)2

= -58.5J
E(net) = E(q1) + E(q2) = -168.75J -58.5J = -227.25J [-ve sign means it points in the negative x- direction]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.