(a) 5172 V/m
The electric field strength of a uniform field between two parallel plates is given by:
[tex]E=\frac{V}{d}[/tex]
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
[tex]E=\frac{3\cdot 10^3}{0.580}=5172 V/m[/tex]
(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:
[tex]U=K[/tex]
[tex]q\Delta V = \frac{1}{2}mv^2[/tex]
where
[tex]q= 1e=1.6\cdot 10^{-19}C[/tex] is the charge of the ion
[tex]\Delta V=3\cdot 10^3 V[/tex] is the potential difference
[tex]m=3.17\cdot 10^{-26} kg[/tex] is the mass of the ion
v is the final speed of the ion
Solving for v, we find
[tex]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[/tex]
(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
[tex]F=m\frac{v^2}{r}[/tex]
where
r is the radius of the circular path
So we can write
[tex]qvB = m\frac{v^2}{r}[/tex]
and solving for the radius, we find
[tex]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[/tex]
