Question

determine the magnitude of an electric field that will balance the weight of an electron.

Answer 1

Answer:

5.6 X 10^-11 NC^-1. is the correct ans

Answer 2

To find:

Magnitude of an electric field that will balance the weight of an electron.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(3,3){\circle{2}}\put(3,3){\vector(0,1){2}}\put(3,3){\vector(0,-1){2}}\put(4,5){ \rm{F_{E}} }\put(4,1){ \rm{F_{g}} }\put(2,3){ \rm{e} }\end{picture}}$

Calculation:

Weight of an electron is balanced only when the electrostatic force experienced by the electron will be equal and opposite to that of the weight of the electron.

$\therefore\: F_{g}=F_{E}$

$=> \: mg = Eq$

$= > 9.1 \times {10}^{ - 31} \times 10 = E \times (1.6 \times {10}^{ - 19} )$

$= > 9.1 \times {10}^{ - 30} = E \times (1.6 \times {10}^{ - 19} )$

$= > 9.1 \times {10}^{ (- 30 + 19)} = E \times (1.6 )$

$= > 9.1 \times {10}^{ - 11} = E \times (1.6 )$

$= >E = \dfrac{ 9.1 \times {10}^{ - 11}}{1.6}$

$= >E = 5.68 \times {10}^{ - 11} \: N {C}^{ - 1}$

So, final answer is:

$\boxed{ \sf{E = 5.68 \times {10}^{ - 11} \: N {C}^{ - 1} }}$