So the question reads: \"A small drop of water is suspended motionless in air by

Question

So the question reads:

"A small drop of water is suspended motionless in air by a uniformelectric field that is directed upward and has a magnitude of 8480N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is theexcess charge on the water drop positive or negative? Why? (b) Howmany excess electrons or protons reside on the drop?"

I know this so far, how do you do part b?(a) Positive, sothat the electrostatic force points upward.
(b) ?
F=qE?? is gravity involved for F=ma?
So the question reads:

"A small drop of water is suspended motionless in air by a uniformelectric field that is directed upward and has a magnitude of 8480N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is theexcess charge on the water drop positive or negative? Why? (b) Howmany excess electrons or protons reside on the drop?"

I know this so far, how do you do part b?(a) Positive, sothat the electrostatic force points upward.
(b) ?
F=qE?? is gravity involved for F=ma?

Explanation / Answer

   For equilibrium,   upwardselectrostatic force   =   downwardsforce i.e. weight    q *E   =   m * g    q   =   3.50 *10-9 * 9.8 /8480   =   4.04 *10-12   C    Since charge isquantized,   q   =   n* e   =>   4.04 *10-12   =   n * 1.6 *10-19    no. of excessprotons   n   =   4.04* 10-12 / 1.6 *10-19   =   2.525 *107   electrons    no. of excessprotons   n   =   4.04* 10-12 / 1.6 *10-19   =   2.525 *107   electrons

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