Answer:
Explanation:
The weight of the bee is:
[tex] W=mg=(120\times10^{-6}kg)(9.81\frac{m}{s^{2}})=1.18\times10^{-3}N[/tex]
with m the mass and g the gravity acceleration.
Electric force of the bee is related with the electric field of earth by:
[tex]F_{e}=qE=(23\times10^{-12}C)(100\frac{N}{C})=2.3\times10^{-9} [/tex]
with q the charge, E the electric field and Fe the electric force.
So:
[tex] \frac{F}{W}=\frac{2.3\times10^{-9}}{1.18\times10^{-3}}=1.95\times10^{-6}[/tex]
Because Newton's first law we should make the net force on it equals cero:
[tex] F+F_e+W=0[/tex]
[tex]F=-(F_e+W)=-(2.3\times10^{-9} +1.18\times10^{-3})=-1.1800023\times10^{-3} [/tex]
with W the weight, Fe the electric force on the bee due earth's electric field and F the force.
So, the applied electric field should be:
[tex]E_a=\frac{-1.1800023\times10^{-3}}{23\times10^{-12}}=-51304447 \frac{N}{C} [/tex]
The negative sign indicates that the electric field should be opposite to earth's electric field, so it should be upward.
