A tiny dust particle of mass of 2.0 times 10^-6 kg and net charge of+1.0 times 1

Question

A tiny dust particle of mass of 2.0 times 10^-6 kg and net charge of+1.0 times 10^-8 C hangs suspended in equilibrium between the horizontal, parallel plates of a capacitor. sigma F = ma What is the acceleration of the stationary dust particle? a = ___ m/s^2 From the figure, you immediately jump to the obvious conclusion that sigma F = F_electric -F_gravity Since the force of gravity could be given by either F_gravity = G middot m middot M_earth/r^2_earth or F_gravity = mg and since you have searched in vain for numerical values of both M_earth, and G, and cannot find them, but you do see on the cover page, and also remember from Ph218. that g=9.8 m/s^2, therefore it becomes immediately obvious which equation you should use. Ta Da! Use your Sherlock Holmes-ian deductive powers to calculate the force of the electric field. What is the force caused by the electric field (in Newtons)? |F| = ___N Show that V/m are same units as same as N/C by breaking down, rearranging, and/or recombining individual units in the ratio. You this may take as many steps as needed. What is the numerical magnitude of the electric field at the location of the dust particle in V/m? |E|=___N/C Note that magnitude does NOT contain a plusminus sign. Draw the vector direction of the ELECTRIC field on the figure next to the appropriate force vector.

Explanation / Answer

32.

mass = m = 2.0x10-6 kg, charge, q = 1.0x10-8 C,

F = mg = qE

Electric field, E = mg/q = 2.0x10-6 *9.8 / 1.0x10-8 = 1.96e+3 N/C

Acceleration, a = g = 9.8 m/s2

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