Answer to Question #315186 in Electric Circuits for Leo

Question

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

a.) an electron

b.) a proton

show your solution

Answer 1

We know that

Electric force =weight

$qE=mg$

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

(a)

For electron

$$E=-\frac{9.1\times10^{-31}\times9.8}{1.6\times10^{-9}}=-5.57\times10^{-11}N/C\\|E|=5.57\times10^{-11}N/C$$

Direction will be downward charge is negative

(B)

For proton

$$E=\frac{1.67\times10^{-27}\times9.8}{1.6\times10^{-19}}=1.022\times10^{-7}N/C$$

direction will be upward charge for positive