Consider the figure above. The blue balls are located at (-1,1,0) and (1,-1,0) e

Question

Consider the figure above. The blue balls are located at (-1,1,0) and (1,-1,0) each carry a charge, Q. The green balls are located at (1,1,0) and (-1,-1,0). They each carry a charge -Q. There are three points marked in red, O is the origin, P1 is (0,0,2) and P2 is (2,0,0). There are no charges at these points. I would like to rank the points O, P1, and P2 in order of decreasing electric field strength. The correct answer is:

A.The fields are all equal
B.P1>O>P2
C.O>P1>P2
D.O>P2>P1
E.P1>P2>O
F.P2>P1=O
G.O>P2=P1
H.P2>P1>O
I.P1=P2>O
J.None of the above

Explanation / Answer

charge at (x0, y0, z0), the point to get field is (x, y, z)

the magnitude of the electric field strenth is

Kq[(x - x0)i + (y - y0) j + (z - z0)k]/[(x - x0)2 + (y - y0)2 + (z - z0)2]3/2

for O: x = y = z = 0

EO/(KQ) = (1i - 1j)/[(1)2 + (-1)2]3/2 + (-1i + 1j)/[(-1)2 + (1)2]3/2 - (-1i - 1j)/[(-1)2 + (-1)2]3/2 - (1i + 1j)/[(1)2 + (1)2]3/2 = 0

for P1: x = y = 0, z = 2

EP1/(KQ) = (1i - 1j + 2k)/[(1)2 + (-1)2 + 22]3/2 + (-1i + 1j + 2k)/[(-1)2 + (1)2 + 22]3/2 - (-1i - 1j + 2k)/[(-1)2 + (-1)2 + 22]3/2 - (1i + 1j + 2k)/[(1)2 + (1)2 + 22]3/2 = 8/63/2 k = 0.544 k

for P2: x = 2, y = z = 0

EP2/(KQ) = (3i - 1j)/[(3)2 + (-1)2]3/2 + (1i + 1j)/[(1)2 + (1)2]3/2 - (1i - 1j)/[(1)2 + (-1)2]3/2 - (3i + 1j)/[(3)2 + (1)2 ]3/2 = 2/23/2 = 1/2 = 0.707 j

answer: H

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