Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
By definition, the potential energy is:
U = qV
Where,
q: load
V: voltage.
Then, the kinetic energy is:
K = mv ^ 2/2
Where,
m: mass
v: speed.
As the power energy is converted into kinetic energy, we have then:
U = K
Equating equations:
qV = mv ^ 2/2
From here, we clear the speed:
v = root (2qV / m)
Substituting values we have:
v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
v = 3.56 × 10 ^ 7 m / s
Then, the centripetal force is:
Fc = Fm
mv ^ 2 / r = qvB
By clearing the magnetic field we have:
B = mv / qr
Substituting values:
B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
B = 3.43 × 10 ^ -3 T
Answer:
A magnetic field that must be experienced by the electron is:
B = 3.43 × 10 ^ -3 T
Answer:
the yellow one
Explanation:
2 of the same elements resolute as the same element
