The mass of the sheet of aluminium is the product between its density d and its volume V:
Re-arranging the equation, we can find the volume of the sheet:
Converted in millimeters,
The area is
Since
, this area corresponds to
So we can now find the thickness of the sheet of aluminium:
Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.
An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?
Answer:
Work=1.06×10⁻²¹J
Explanation:
Given Data
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Charge density σ=3.90×10⁻¹² C/m²
The electron moves a distance d=3.00×10⁻²m
Electron charge e=-1.6×10⁻¹⁹C
To find
Work done
Solution
The electric field due is sheet is given as
E=σ/2ε₀
Now we need to find force on electron
Now for Work done on the electron
False.
Unweathered rock that underlies soul is known as bedrock.
