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?,

Question

Physics

Sadie Williams

24 May, 20:19

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?,

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Alan Grant · Answer 1

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