Complete Question
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,
A :
a force repels the glass out of the capacitor.
B :
a force attracts the glass into the capacitor.
C :
no force acts on the glass.
D :
a net charge appears on the glass.
E :
the glass makes the plates repel each other.
Answer:
The correct option is B
Explanation:
Generally when the glass dielectric is slowly inserted between the plated,
The positive plate of the capacitor will induce a negative charge on the glass while the negative plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass
Answer:
D
Explanation:
First we define our variables
V0=29.4
a=-9.8
V=0
We have to find the maximum displacement , which I will define as X
We use formula v^2=v0^2+2aX
All we do is substitute our values
0=29.4^2-19.6X
29.4^2=19.6X
X=29.4^2/19.6=44.1
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Answer:
fact one
Explanation:
fact two is going slower and a longer distance than fact one so fact one will get there first.
hope this helps
