A tiny oil drop of mass, 0.0000001kg of mass and -1.6 * 10-10C of charge is sitt

Question

A tiny oil drop of mass, 0.0000001kg of mass and -1.6 * 10-10C of charge is sitting in the vacuum under the influence of earth's gravity, g = 9.8ms-2. Calculate the magnitude of the electric field intensity, E in units of NC-1 required to balance out the oil drop in vacuum preventing the falling down due to gravity.

In 1909, Robert Millikan (Links to an external site.)Links to an external site. used this simple idea to measure the electron's charge thus earning the Nobel prize for physics.  

WHICH MY ANSWER I GOT IS 6125, BUT MY QUESTION IS THIS PART THAT IS CONFUSING ME.

In above question, what direction should the electric field, E be applied?

A. Vertically Up

B. Toward East

C. Toward West

D. Toward South

E. Vertically Down

Explanation / Answer

Given

Charge Q = - 1.6 x 10^-10 C

Mass m = 0.0000001 kg

Force, F = QE = mg

=> Electric field E = mg/Q = (0.0000001 x 9.8) /(-1.6 x 10^-10)

Electric field E = 6125 N/C  

Direction will be vertically downward

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