Question
The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $$9.1 \times {10^{ - 31}}kg$$ and $$ - 1.6 \times {10^{ - 19}}C.$$ )
A.
$$ - 5.6 \times {10^{ - 11}}N/C$$
B.
$$ - 4.8 \times {10^{ - 15}}N/C$$
C.
$$ - 1.6 \times {10^{ - 19}}N/C$$
D.
$$ - 3.2 \times {10^{ - 19}}N/C$$
Answer :
$$ - 5.6 \times {10^{ - 11}}N/C$$
Solution :
$$\eqalign{
& - eE = mg \cr
& \vec E = - \frac{{9.1 \times {{10}^{ - 31}} \times 10}}{{1.6 \times {{10}^{ - 19}}}} = - 5.6 \times {10^{ - 11}}\,N/C \cr} $$