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A very small object with mass 6.90 × 10^−9 kg and positive charge 8.30 × 10^−9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 x 10^-8 C/m^2. The object is initially 0.530 m from the sheet.

What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.170 m ?,

Respuesta :

Kalahira
Mass of the object m = 6.90 × 10^−9 kg Charge of the object Q = 8.30 × 10^−9 C Surface charge density = 5.90 x 10^-8 C/m^2 Distance d1 = 0.530 m, d2 = 0.170 m Kinetic energy KE = 1/2 mv^2 Work = F x d = electric field x q x d => E x 8.30 × 10^−9 x (0.530 - 0.170) Following the energy conversion principles, 1/2 x 6.90 × 10^−9 x v^2 = E x 8.30 × 10^−9 x (0.530 - 0.170) 3.45 x v^2 = 8.30 x 0.36 x E => 3.45v^2 = 3E calculating the electric field E = 5.90 x 10^-8 / ( 2 x 8.85 x 10^-12) = 3333 So 3.45v^2 = 3 x 3333 => v^2 = 9999 / 3.45 => v^2 = 2898 => v = 53.8 m/s Intial speed v = 53.8 m/s

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