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CBSE Class 12-science Answered

How many electrons should be removed from a coin of mass 3.2 g, so that it just float in an electric field of intensity 1010 NC-1, directed upward.
Asked by Topperlearning User | 17 Apr, 2015, 12:46: PM
answered-by-expert Expert Answer

Here, m = 3.2g = 3.2  10-3 kg

E = 1010 NC-1
Let n be the number of electrons removed from the coin. Then the charge on the coin is,
q = +ne
 When the coin just floats,
Upward force of electric field = weight of the coin
nqE = mg
 
begin mathsize 11px style straight n space equals space mg over qE space equals space fraction numerator 3.2 space cross times space 10 to the power of negative 3 end exponent space cross times space 9.8 over denominator 1.6 space cross times space 10 to the power of negative 19 end exponent space cross times space 10 to the power of 10 end fraction space equals space 1.96 space cross times space 10 to the power of 7 space electrons end style 
Answered by | 17 Apr, 2015, 02:46: PM
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