Answer:
v = √ 2e (V₂-V₁) / m
Explanation:
For this exercise we can use the conservation of the energy of the electron
At the highest point. Resting on the top plate
Em₀ = U = -e V₁
At the lowest point. Just before touching the bottom plate
Emf = K + U = ½ m v² - e V₂
Energy is conserved
Em₀ = Emf
-eV₁ = ½ m v² - e V₂
v = √ 2e (V₂-V₁) / m
Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron
Answer:
Explanation:
Force on a moving charge is given by the following relation
F = q ( v x B )
for proton
q = e , v = vi , B = Bk
F = e ( vi x Bk )
= Bev - j
= - Bevj
The direction of force is along negative of y axis or -y - axis.
for electron
q = - e , v = vi , B = Bk
F = - e ( vi x Bk )
= - Bev - j
= Bevj
The direction of force is along positive of y axis or + y - axis.
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
