You have a great summer job in a research laboratory with a group investigating the possibility of producing power from fusion. The device being designed confines a hot gas of positively charged ions, called plasma, in a very long cylinder with a radius of 2.0 cm. The charge density of the plasma in the cylinder is 6.0 x 10-5 C/m3. Positively charged Tritium ions are to be injected into the plasma perpendicular to the axis of the cylinder in a direction toward the center of the cylinder. Your job is to determine the speed that a Tritium ion should have when it enters the plasma cylinder so that its velocity is zero when it reaches the axis of the cylinder. Tritium is an isotope of Hydrogen with one proton and two neutrons. You look up the charge of a proton and mass of the tritium in your trusty Physics text to be 1.6 x 10-19 C and 5.0 x 10-27 Kg.

Generally the difference in potential energy of the Tritium ion before entering the cylinder and at the axis of the cylinder is mathematically represented as

[tex]\Delta U= U_f - U_i = \frac{Q * J * r^2}{ 4 * \epsilon_o}[/tex]

Here

[tex] \epsilon_o}[/tex] is the permitivity of free space with value

[tex] \epsilon_o} =8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

So

[tex]\Delta U = \frac{1.60 *10^{-19} * 6.0 * 10^{-5} * 0.02^2}{ 4 * 8.85*10^{-12}}[/tex]

[tex]\Delta U = 1.085 *10^{-16} \ J [/tex]

Generally from energy conservation we have that

[tex]KE_i + U_i = KE_f + U_f[/tex]

Given that the velocity at the axis of the cylinder should be zero then

it means that the kinetic energy at this axis will also be zero

So

[tex]KE_i + U_i = 0 + U_f[/tex]

Here KE_i is the initial kinetic energy of the Tritium ion which is mathematically represented as

[tex]KE_i = \frac{1}{2} * m * v^2[/tex]

So

=> [tex] \frac{1}{2} * m * v^2 = U_f - U_i[/tex]

=> [tex] \frac{1}{2} * m * v^2 = \Delta U[/tex]

=> [tex] \frac{1}{2} * m * v^2 = 1.085 *10^{-16}[/tex]

=> [tex]v =\sqrt{\frac{ 1.085 *10^{-16}}{0.5 * m} }[/tex]

Here m is the mass of Tritium ion which is given in the question

=> [tex]v =\sqrt{\frac{ 1.085 *10^{-16}}{0.5 * 5.0 x 10^{-27} } }[/tex]

=> [tex]v = 2.083 *10^{5} \ m/s[/tex]