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A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d = 35cm above charge Q1. The net force on Q2 is equal to zero.
A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d = 35cm above charge Q1. The net force on Q2 is equal to zero.a) Write an equation for the magnitude of charge Q2 in terms of the given variables.
b) Calculate the magnitude of Q2 in units of nanocoulombs
The charge Q₂ is(a) in terms of variables expressed as(b) The magnitude of charge Q₂ = 0.01334 nCGiven two charges Q₁ and Q₂ such thatthe charge on Q₁ = 35 nC = 35×10⁻⁹ CMass of Q₂ is m = 3.5 microgram = 3.5×10⁻⁹ Kgdistance r between the charges = 35 cm = 0.35 m(a)The charge Q₂ is floating because theelectrostatic forceF and the weight W of the charge are balanced, that isF = W(b)The magnitude of Q₂:Q₂ = 13.34×10⁻¹² CQ₂ = 0.01334 nC must be the magnitude of the charge so that it floats.Learn more:brainly.com/question/14870624...
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