20075ee1_1_HW2

# 20075ee1_1_HW2 - CHAPTER 2 2.1 Four 10nC positive charges...

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CHAPTER 2 2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for = 0 : Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge will be on the z axis at location z = 4 2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be z -directed, and will be four times the z component of force produced by each of the four other charges. F = 4 2 × q 2 4 π 0 d 2 = 4 2 × (10 8 ) 2 4 π (8 . 85 × 10 12 )(0 . 08) 2 = 4 . 0 × 10 4 N 2.2. Two point charges of Q 1 coulombs each are located at (0,0,1) and (0,0,-1). (a) Determine the locus of the possible positions of a third charge Q 2 where Q 2 may be any positive or negative value, such that the total field E = 0 at (0,1,0): The total field at (0,1,0) from the two Q 1 charges (where both are positive) will be E 1 (0 , 1 , 0) = 2 Q 1 4 π 0 R 2 cos45 a y = Q 1 4 2 π 0 a y where R = 2. To cancel this field, Q 2 must be placed on the y axis at positions y > 1 if Q 2 > 0, and at positions y < 1 if Q 2 < 0. In either case the field from Q 2 will be E 2 (0 , 1 , 0) = −| Q 2 | 4 π 0 a y and the total field is then E t = E 1 + E 2 = Q 1 4 2 π 0 | Q 2 | 4 π 0 = 0 Therefore Q 1 2 = | Q 2 | ( y 1) 2 y = 1 ± 2 1 / 4 | Q 2 | Q 1 where the plus sign is used if Q 2 > 0, and the minus sign is used if Q 2 < 0. (b) What is the locus if the two original charges are Q 1 and Q 1 ? In this case the total field at (0,1,0) is E 1 (0 , 1 , 0) = Q 1 / (4 2 π 0 ) a z , where the positive Q 1 is located at the positive z (= 1) value. We now need Q 2 to lie along the line x = 0, y = 1 in order to cancel the field from the positive and negative Q 1 charges. Assuming Q 2 is located at (0 , 1 , z ), the total field is now E t = E 1 + E 2 = Q 1 4 2 π 0 a z + | Q 2 | 4 π 0 z 2 = 0 or z = ± 2 1 / 4 | Q 2 | /Q 1 , where the plus sign is used if Q 2 < 0, and the minus sign if Q 2 > 0. 1

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2.3. Point charges of 50nC each are located at A (1 , 0 , 0), B ( 1 , 0 , 0), C (0 , 1 , 0), and D (0 , 1 , 0) in free space. Find the total force on the charge at A . The force will be: F = (50 × 10 9 ) 2 4 π 0 R CA | R CA | 3 + R DA | R DA | 3 + R BA | R BA | 3 where R CA = a x a y , R DA = a x + a y , and R BA = 2 a x . The magnitudes are | R CA | = | R DA | = 2, and | R BA | = 2. Substituting these leads to F = (50 × 10 9 ) 2 4 π 0 1 2 2 + 1 2 2 + 2 8 a x = 21 . 5 a x µ N where distances are in meters.
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