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 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.

Answer 1

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$