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
2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley
Answer:
W = 5586[J]
Explanation:
Work in physics is defined as the product of force by a distance.

where:
W = work [J]
F = force = m*g [N]
m = mass = 300 [kg]
g = gravity acceleration = 9.81 [m/s²]
d = distance = 1.9 [m]
Now replacing:
![W=m*g*d\\W=300*9.81*1.9\\W=5586[J]](https://tex.z-dn.net/?f=W%3Dm%2Ag%2Ad%5C%5CW%3D300%2A9.81%2A1.9%5C%5CW%3D5586%5BJ%5D)
Answer:
Explanation:
capacitance of each capacitor
C₀= Q₀ / V₀
V₀ = Q₀ / C₀
New total capacitance = C₀ ( 1 + K )
Common potential
= total charge / total capacitance
= 2 Q₀ / [ C₀ ( 1 + K ) ]
2 V₀ / ( 1 + K )
b )
Common potential = 2 x V₀ / ( 1 + 7.8 )
= .227 V₀
charge on capacitor with dielectric
= .227 V₀ x 7.8 C₀
= 1.77 V₀C₀
= 1.77 Q₀
Ratio required = 1.77