18_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

18_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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18 CHAPTER 15 15.26 If the resultant field is to be zero, the contributions of the two charges must be equal in magnitude and must have opposite directions. This is only possible at a point on the line between the two negative charges. Assume the point of interest is located on the y -axis at . Then, for equal magnitudes, 4.0 m 6.0 m y −< < 12 22 ee kq rr = or () C 8.0 C m 4.0 m yy 9.0 6.0 µ = −+ Solving for y gives 8 6.0 m 9 += 4.0 m , or 0.85 m y =+ ± ± ²³´µ¶ ± · ² ± ¸ ³ ± ³ ¸ ¹³´µ¶ ´ ± µºµ»¼³´µ ± ½ ´ ¸ µºµ»¾³´µ ± ½ ±² 15.27 If the resultant field is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x -axis, with the origin at the –2.5 C charge. Then, the two contributions will have opposite
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