Please answer Part (a) For i and part (b) for i as well www.webassign.net/web/st

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Please answer Part (a) For i
and part (b) for i as well

www.webassign.net/web/student/Assignment.R 7. 0.5/1 points | Previous Answers SerPSET9 23 P031 Three point charges are located on a circular arc as shown in the figure below. (Taker 4.12 cm Let to the right be the +x direction and up along the screen be the ty direction.) in the figure below. (Take r = 4.12 cm. Let to the right be the +x d +3.00 nC -2.00 nC 30.0° +3.00 nC (a) What is the total electric field at P, the center of the arc? Yourresponse iffers from the correct answer by more than 10%. Double check your calculations. (b) Find the electric force that would be exerted on a 4.72-nC point charge placed at P Two equal positively charged particles are at opposite corners of a trapezoid as shown in the figure below. (Use the following as necessary: e, d, ke)

Explanation / Answer

Given,

q1 (the one on top) = +3.00 nC

q2 (the one in the middle) = -2.00 nC

q3 (the one at the bottom) = +3.00 nC

Thus, using the equation

E = kq/r^2,

where k = 8.99E9 Nm^2/C^2, q = the source charge, r = the distance of that reference point from the charge.

Thus, the magnitude of the electric field for each is, as r = 4.12 cm = 0.0412 m

E1 = 1.59E4 N/C [towards quadrant 4]

E2 = -1.06E4 N/C [towards the -x axis]

E3 = 1.59E4 N/C [towards quadrant 1]

Now, we get the x and y components of these,

E1x = 13769.80 N/C

E1y = -7950 N/C

E2x = -1.06E4 N/C

E2y = 0 N/C

E3x = +13769.80 N/C

E3y = +7950 N/C

Thus, adding x and y components

Enetx = (13769.80 - 1.06E4+ 13769.80) N/C = 16939.6 N/C

Enety = (-7950 + 0 + 7950) N/C = 0 N/C

Thus, as you see, the net electic field is purely along x.

Thus, the answer to part a) is

Part a: 1.69E4 N/C, to the right

For part b)

F = Eq

Thus, as q = 4.72 nC = 4.72E-9 C

F = 16939.6 N/C * 4.72E-9 C

F = 8.00E-5 N [answer for part b]

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