Question

2. A small piece of dust is located in between two oppositely charged parallel plates where there exists a uniform electric field of magnitude E = 5000N/C. The dust is made up of 1x1015 protons (1 Quadrillion). The distance between the plates is 5cm. a) What is the total charge on the dust particle? b) If the charge moves from the positive plate to the negative plate, what is its change in potential energy? c) We’ve often used the analogy of mass and gravity to understand charge and electric fields. Suppose you placed a bead of mass 0.05kg (50g) on an inclined plane. How tall should the inclined plane be if the bead is to experience the same change in potential energy as the charge between the plates? Give your answer in centimeters

Answer 1

a

The total charge is
`q = 1.6*10^(-4)C`

b

The change in potential energy is
`\Delta U = -4*10^(-2)J`

c

The height is
`h= 8.2cm`

Step-by-step explanation:

From the question we are told that the

The magnitude of electric field is
`E = 5000N/C`

The number of proton is
`N_p = 1*10^(15)`

The distance between the plates is
`d = 5cm = (5)/(100) = 0.05m`

The total charge can be mathematically represented as

`q = N_p * e`

Where e is the charge on one proton which has a value of
`=1.6*10^(-19)C`

Substituting values

`q = 1 *10^(15) * 1.6*10^(-19) = 1.6 *10^(-4)C`

The change in potential energy is mathematically represented ads

`\Delta U = -(qE)d`

where the negative sign shows that the work done by the electric force is against the electric field

Substituting values

`\Delta U = - 1.6*10^(-4) * 5000 * 0.05`

`= -4*10^(-2)J`

The mass of the bead is given as 0.05kg

The change in potential due to gravity is mathematically given as

`\Delta U = -mgh`

the negative sign is due to the fact that the height is decreasing

And
`g =9.8m/s^2`

Making h the subject

`h = (\Delta U)/(mg)`

Substituting values

`h = (4*10^(-2))/(0.05 * 9.8)`

`=0.081m = 0.082 *100 = 8.2cm`