Question

Problem 21-40a:

The electric field near the surface of Earth points downward and has a magnitude of 153 N/C. What is the ratio of the magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron?
Answer: 2.74×1012

Problem 21-40b:
What charge should be placed on a plastic ball of mass 5.25 g so that the electric force balances the weight of the ball near Earth's surface?

Answer 1

Part a)

`(F_e)/(F_g) = 2.74 * 10^(12)`

Part b)

`q = 3.37 * 10^(-4) C`

Step-by-step explanation:

As we know that electric force on electric charge is given as

`F = qE`

here we have

`q = 1.6 * 10^(-19)C`

E = 153 N/C

now force is given as

`F = (1.6 * 10^(-19))(153) = 2.45 * 10^(-17) N`

Gravitational force on electric charge near surface of earth is given as

`F_g = mg`

`F_g = (9.1 * 10^(-31))(9.81) = 8.93 * 10^[-30} N`

now the ratio of two forces is given as

`(F_e)/(F_g) = (2.45 * 10^(-17))/(8.93 * 10^(-30))`

`(F_e)/(F_g) = 2.74 * 10^(12)`

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

`F_g = qE`

`mg = qE`

`(5.25 * 10^(-3))(9.81) = q(153)`

here we have

`q = 3.37 * 10^(-4) C`