There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather conditions, in an open field, the strength of this electric field is 95.0 N/C . A spherical pollen grain with a radius of 12.0 μm is released from its parent plant by a light breeze, giving it a net charge of −0.700 fC (where 1 fC=1×10−15 C ). What is the ratio of the magnitudes of the electric force to the gravitational force, ????electric/????grav , acting on the pollen? Pollen is primarily water, so assume that its volume mass density is 1000 kg/m3 , identical to the volume mass density of water.

Respuesta :

Answer:

[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]

Explanation:

Radius of the pollen is given as

[tex]r = 12.0 \mu m[/tex]

Volume of the pollen is given as

[tex]V = \frac{4}{3}\pi r^3[/tex]

[tex]V = \frac{4}{3}\pi (12\mu m)^3[/tex]

[tex]V = 7.24 \times 10^{-15} m^3[/tex]

mass of the pollen is given as

[tex]m = \rho V[/tex]

[tex]m = 7.24 \times 10^{-12}[/tex]

so weight of the pollen is given as

[tex]W = mg[/tex]

[tex]W = (7.24 \times 10^{-12})(9.81)[/tex]

[tex]W = 7.1 \times 10^{-11}[/tex]

Now electric force on the pollen is given

[tex]F = qE[/tex]

[tex]F = (-0.700\times 10^{-15})(95)[/tex]

[tex]F = 6.65 \times 10^{-14} N[/tex]

now ratio of electric force and weight is given as

[tex]\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}[/tex]

[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]