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trasher [3.6K]
3 years ago
13

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.
Physics
1 answer:
lozanna [386]3 years ago
3 0

Answer:

The value is v = 2.083 *10^{5} \  m/s

Explanation:

From the question we are told that

The radius of the cylinder is r =  2.0 \ cm  =  0.02 \  m

The charge density of the plasma is J  =  6.0 * 10^{-5} C/m3

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

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

Here

\epsilon_o} is the permitivity of free space with value

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

So

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

\Delta U  =  1.085 *10^{-16} \  J

Generally from energy conservation we have that

KE_i + U_i = KE_f + U_f

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

KE_i + U_i = 0 + U_f

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

KE_i  =  \frac{1}{2} *  m * v^2

So

=> \frac{1}{2} *  m * v^2 =  U_f  - U_i

=> \frac{1}{2} *  m * v^2 = \Delta U

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

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

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

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

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

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Answer:

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Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

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F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
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Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

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q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

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