Question

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 110.0 N/C 110.0 N/C . A spherical pollen grain with a radius of 11.0 μm 11.0 μm is released from its parent plant by a light breeze, giving it a net charge of − 0.550 fC −0.550 fC (where 1 fC = 1 × 10 − 15 C 1 fC=1×10−15 C ). What is the ratio of the magnitudes of the electric force to the gravitational force, F e l e c t r i c / F g r a v Felectric/Fgrav , acting on the pollen? Pollen is primarily water, so assume that its volume mass density is 1000 k g / m 3 1000 kg/m3 , identical to the volume mass density of water.

Answer 1

Answer:

0.00110616446408

Explanation:

r =  Radius of sphere = 11 μm

V = Volume = $\dfrac{4}{3}\pi r^3$

$\rho$ = Density of pollen = 1000 kg/m³

q = Charge = − 0.550 fC

E = Electric field = 110 N/C

g = Acceleration due to gravity = 9.81 m/s²

Mass is given by

$m=\rho V\\\Rightarrow m=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\ kg$

Force due to gravity is given by

$F_g=mg\\\Rightarrow F_g=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\times 9.81\\\Rightarrow F_g=5.4693494471\times 10^{-11}\ N$

Electric force is given by

$F_e=qE\\\Rightarrow F_e=-0.55\times 10^{-15}\times 110\\\Rightarrow F_e=-6.05\times 10^{-14}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{6.05\times 10^{-14}}{5.4693494471\times 10^{-11}}\\\Rightarrow \dfrac{F_e}{F_g}=0.00110616446408$

The ratio is 0.00110616446408