the magnitude of a vertical electric field which can just suspend an alpha particle of mass 6.4*10^-27 kg freely in air
Answers
Answer:
The electric field balancing the alpha particle is 2x10^-7
Explanation:
The formula for a force on a charged particle in an electric field of strength E is
Force = charge x electric field
Now we need a similar force being experienced by the alpha particle;
F=mg
Hence we can write;
Q x E=m x g
E = m x g /q
As per given data;
Mass = 6.4 x10^-27 kg
G = 10ms–²
Charge on alpha particle = 3.2×10^-19 Coulomb
putting the values in the equation,
The electric field balancing the alpha particle is 2x10^-7
The magnitude of a vertical electric field which can just suspend an alpha particle of mass freely in air is 4 × 10⁻⁷ N/C.
Explanation:
The force acting on the alpha particle is given by the formula:
Force = Charge × Electric field
The force is given by the formula:
m × g = Charge × Electric field
Now,
Electric field = (m × g)/Charge
Where,
m = Mass of alpha particle = 6.4 × 10⁻²⁷ kg
g = Acceleration due to gravity = 10 m/s²
Charge of alpha particle = 1.6 × 10⁻¹⁹ C
On substituting the values, we get,
Electric field = (6.4 × 10⁻²⁷ × 10)/(1.6 × 10⁻¹⁹)
∴ Electric field = 4 × 10⁻⁷ N/C