Electrical field intensity (Grade 12 physics)?

Question

"Using a spectrometer, students accelerate a singly charged ion through a potential difference
of 95.0 V. It then enters a uniform magnetic field of strength 1.12 x 10– 1 T at right angles to its motion. The students measure the radius of curvature of the ion in the magnetic field and find it to be 6.60 x 10– 2 m. Using appropriate equations and logical method, find the mass of this ion."


Basically I know that

Brq/v = m
B, r, and q, are given in the question. My question is how can I find the velocity.

to find v, I assume we can do v = E/B, and E is can be found with E = V/d (Voltage and distance between plates), we are given V, however there is no d to use.

Am I missing something or am I just going about solving this all wrong?

Answer 1

After the ion has been accelerated by the 95 V potential difference, it's potential energy (in electron volts) has decreased by 95 V* |q| , where |q| is the magnitude of the charge on the ion ( equal to 1 in this case because we are told the ion is singly charged). That decrease in potential energy must be balanced by an equivalent increase in the ion's kinetic energy. You know that for velocities much less than the speed of light in a vacuum, KE = 0.5*m*v^2, where v is the speed and m is the mass.



1 electron volt (eV) = 1.602* 10^-19 J, so 95 eV = 1.522*10^-17 J, which is the kinetic energy of the ion after being accelerated.



You now have a new equation involving the two unknowns, v and m. Combine this with the equation involving the radius of curvature and the magnetic field strength to eliminate v between the equations, and you can then solve for m.

Answer 2

The strength of the electric field is measured using a quantity called the electric field intensity. The greater the electrical field intensity the stronger the field. The electrical field intensity (EE) is defined as: The electric field intensity is the force on a unit positive charge placed at that point in the field.

In a uniform field the electric field intensity is constant (the same at any point in the field) while in a radial field the electric field intensity decreases as the distance from the central charge increases.



Therefore for a radial field the electric field intensity distance d from a positive charge of size Q coulombs is:



Radial field: EE = (1/4πεo)Q/d2



For a uniform field between two parallel plates separated by a distance d and with a potential difference V between them the field is:



Uniform field: EE = V/d



the units for electric field intensity are Newtons per coulomb (NC-1).

Answer 3

Electric field intensity is defined as the strength of an electric field at any point. It is equal to the force per unit charge experienced by a test charge placed at that point. The unit of measurement is the volt/meter.

Answer 4

The question probably needs you to ignor the gravitational container from the solar. enable the '0' factor be a distance (De) from Earth and (Dm) from Moon. using a 1kg 'attempt' mass at this factor, and thinking the gravitational forces (Fg) in this mass, from Earth and Moon, as equivalent .. Me = mass of Earth (5.ninety 8*10^24 kg) Mm = mass of Moon (7.35*10^22 kg) Fg = G(Me*a million)/De² = G(Mm*a million)/Dm² ........... Me/De² = Mm/Dm² (De/Dm) = ?( Me/Mm) = ? {(5.ninety 8*10^24)/(7.35*10^22 )} ? 9.0 (9.02) • De = 9 Dm ... ie. De = 9/(9+a million) = 0.9 of distance Earth to Moon recommend Earth to Moon distance = 3.80 4*10^8 m .... De = 0.9*(3.80 4*10^8 ) .. ?De = 3.456*10^8 m