show work and answer right if not answer will be given to someone else In 1909 R
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show work and answer right if not answer will be given to someone else
In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that, the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 times 10-19 C - the charge of the electron. For this he won the Nobel Prize. If a drop of mass 1.63265 times 10-13 kg remains stationary in an electric field of 5 times 105 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s2. Answer in units of C How many extra electrons are on this particular oil drop (given the presently known charge of the electron)?Explanation / Answer
q*E=m*g (by force balance) q*5*10^5=1.63265*10^-13 *9.8 q=3.1999*10^-18 C
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