Question

A positively charged ion with charge equivalent to 1e and mass 3.17 x 10^−26 kg is accelerated in a vacuum between two parallel plates with a potential difference of 3.00 x10^3 V between them. The plates are separated by a distance of 0.580 m.

(a) What is the electric field strength between the plates?
(b) Determine its speed when it reaches the far plate.
(c) If the ion then moves into a uniform magnetic field of 5.19 x 10^−2 T, determine the
radius of its path as it is deflected by the magnetic field.

Answer 1

(a) 5172 V/m

The electric field strength of a uniform field between two parallel plates is given by:

$E=\frac{V}{d}$

where

V is the potential difference between the plates

d is the distance between the plates

Here we have

V = 3.00 x 10^3 V

d = 0.580 m

So the electric field strength is

$E=\frac{3\cdot 10^3}{0.580}=5172 V/m$

(b) 1.74 x 10^5 m/s

According to the law of conservation of energy, the electric potential energy of the ion is converted into kinetic energy. Therefore we can write:

$U=K$

$q\Delta V = \frac{1}{2}mv^2$

where

$q= 1e=1.6\cdot 10^{-19}C$ is the charge of the ion

$\Delta V=3\cdot 10^3 V$ is the potential difference

$m=3.17\cdot 10^{-26} kg$ is the mass of the ion

v is the final speed of the ion

Solving for v, we find

$v=\sqrt{\frac{2q\Delta V}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})(3000)}{3.17\cdot 10^{-26}}}=1.74\cdot 10^5 m/s$

(c) 0.664 m

When the ion moves into the magnetic field region, the magnetic force acting on it acts as centripetal force.

The magnetic force is:

F = qvB

B = 5.19 x 10^−2 T is the magnetic field strength

v is the speed of the ion, found in the previous part

While the centripetal force is

$F=m\frac{v^2}{r}$

where

r is the radius of the circular path

So we can write

$qvB = m\frac{v^2}{r}$

and solving for the radius, we find

$r=\frac{mv}{qB}=\frac{(3.17\cdot 10^{-26})(1.74\cdot 10^5)}{(1.6\cdot 10^{-19})(5.19\cdot 10^{-2})}=0.664 m$