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

`(F)/(W) = 9.37 * 10^(-4)`

Step-by-step explanation:

Radius of the pollen is given as

`r = 12.0 \mu m`

Volume of the pollen is given as

`V = (4)/(3)\pi r^3`

`V = (4)/(3)\pi (12\mu m)^3`

`V = 7.24 * 10^(-15) m^3`

mass of the pollen is given as

`m = \rho V`

`m = 7.24 * 10^(-12)`

so weight of the pollen is given as

`W = mg`

`W = (7.24 * 10^(-12))(9.81)`

`W = 7.1 * 10^(-11)`

Now electric force on the pollen is given

`F = qE`

`F = (-0.700* 10^(-15))(95)`

`F = 6.65 * 10^(-14) N`

now ratio of electric force and weight is given as

`(F)/(W) = (6.65 * 10^(-14))/(7.1 * 10^(-11))`

`(F)/(W) = 9.37 * 10^(-4)`