A room with 3.0-m high ceilings has a metal plate on the floorwith V=0V and a se

Question

A room with 3.0-m high ceilings has a metal plate on the floorwith V=0V and a separate metal plate on the ceiling. A 1.0 gglass ball charge to 4.9 nC is shot straight up at 5.0 m/s. How high does the ball go if the ceiling voltage is (a) 3.0*10^6 Vand (b) -3.0*10^6? Physics

Explanation / Answer

a) W = F(net)x = 0 - 0.5mv^2, F(net) = F(coulomb) – mg Note: the final velocity must be zero. Therefore, E(final)=0 F = qE = qV/d = (5.1x10^-9)(3.4x10^6)/3.2 = 5.42x10^-3 N F(net) = F(coulomb:repulsion) – mg = (-5.42x10^-3) – (1.3x10^-3)(9.8) = (-18.2x10^-3) N W = Fh = 0.5mv^2 => h = (0.5mv^2)/F = [0.5(1.3x10^-3)(5.1)^2]/18.2.42x10^-3 = 0.93 m b) F = qE = qV/d = (5.1x10^-9)(2.8x10^6)/3.2 = 4.46x10^-3 N F(net) = F(coulomb:attraction) – mg = 4.46x10^-3 – (1.3x10^-3)9.8 = -8.28x10^-3 N h = 0.5mv^2/F = [0.5(1.3x10^-3)(5.1)^2]/8.28x10^-3 = 2 m

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