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 15.0 μm is released from its parent plant by a light breeze, giving it a net charge of −0.800 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

The ratio of the electric force and the gravitational force is 1.85

Step-by-step explanation:

Given:

• The strength of the electric Field E=95 N/C

• Charge of the pollen grain
  `Q=15*10^(-15) \ C`
• Density of the pollen grain,
  `\rho=1000\ \rm kg/m^3`

Let m be the mass of the pollen grain given by

`m=\rho *(4\pi R^3)/(3)\\m=1000 *(4\pi (15*10^(-6))^3)/(3)\\\\m=4.18*10^(-15)\ \rm kg`

Gravitational force acting on the pollen grains

`=mg\\=4.18*10^(-15)*9.8\\=40.96 *10^(-15)\ N`

The Electric force acting on the pollen grains

`=qE\\=.8*10^(-15)*95\\=76*10^(-15)\ N`

The ratio of electric force to the gravitational force

=
`=(76*10^(-15))/(40*10^(-15))\\=1.855`