Question #315186

What must be the magnitude and direction of E that will balance the weight of

a.) an electron

b.) a proton

show your solution


1
Expert's answer
2022-03-21T12:44:22-0400

We know that

Electric force =weight

qE=mgqE=mg

E=mgqE=\frac{mg}{q}

(a)

For electron


E=9.1×1031×9.81.6×109=5.57×1011N/CE=5.57×1011N/CE=-\frac{9.1\times10^{-31}\times9.8}{1.6\times10^{-9}}=-5.57\times10^{-11}N/C\\|E|=5.57\times10^{-11}N/C

Direction will be downward charge is negative

(B)

For proton


E=1.67×1027×9.81.6×1019=1.022×107N/CE=\frac{1.67\times10^{-27}\times9.8}{1.6\times10^{-19}}=1.022\times10^{-7}N/C

direction will be upward charge for positive


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