Deriving the Voltage Equation: Using Kinematics to Show V=(mv^2/2q)

[Mango12]

Knowing that V=E(delta x), show that V=(mv^2/2q)

I think I have to use kinematics and substitute some things but I'm not sure

 

[topsquark]

Knowing that V=E(delta x), show that V=(mv^2/2q)

I think I have to use kinematics and substitute some things but I'm not sure

There is a lot more to the question than you posted. Do you have a particle in a mass-selector or velocity selector? I'm trying to figure out where the kinetic energy term comes from in V=(mv^2/2q).

-Dan

 

[Mango12]

There is a lot more to the question than you posted. Do you have a particle in a mass-selector or velocity selector? I'm trying to figure out where the kinetic energy term comes from in V=(mv^2/2q).

-Dan

Well, the problem involves xenon. One mole of xenon is .131kg and one atom is 2.2*10^-25 kg. And velocity is 2.7*10^4 m/s if that helps.

He wants us to show that V=(mv^2/2q) is true, so I don't know if he wants us to set them equal to each other or...how else would I prove it is true?

 

[topsquark]

Well, the problem involves xenon.

Please post the whole question, not just what it involves. You still haven't given enough information!

-Dan

 

[Mango12]

Please post the whole question, not just what it involves. You still haven't given enough information!

-Dan

One mole of xenon is .131kg and a single atom of xenon is 2.2*10^-25 kg. It is traveling 2.7*10^4 m/s. Using the diagram and knowing that voltage=E(delta x), show that voltage=mv^2/2q

View attachment 5411

 

[I like Serena]

One mole of xenon is .131kg and a single atom of xenon is 2.2*10^-25 kg. It is traveling 2.7*10^4 m/s. Using the diagram and knowing that voltage=E(delta x), show that voltage=mv^2/2q

Hi Mango12,

When a particle of charge $q$ is accelerated by an electric field $E$, the electric force is $F=qE$.
When traversing a distance $\Delta x$ the work done is $W = F\Delta x$ (if the force is constant).
That work is equal to the kinetic energy $U_k = \frac 12 m v^2$ that the particle gains (assuming it starts from a speed of about zero).

So we have:
$$W = U_k \quad\Rightarrow\quad F\Delta x = \frac 12 m v^2
\quad\Rightarrow\quad qE\Delta x =\frac 12 m v^2
\quad\Rightarrow\quad qV =\frac 12 m v^2
\quad\Rightarrow\quad V =\frac {m v^2}{2q}
$$

 

[Mango12]

Hi Mango12,

When a particle of charge $q$ is accelerated by an electric field $E$, the electric force is $F=qE$.
When traversing a distance $\Delta x$ the work done is $W = F\Delta x$ (if the force is constant).
That work is equal to the kinetic energy $U_k = \frac 12 m v^2$ that the particle gains (assuming it starts from a speed of about zero).

So we have:
$$W = U_k \quad\Rightarrow\quad F\Delta x = \frac 12 m v^2
\quad\Rightarrow\quad qE\Delta x =\frac 12 m v^2
\quad\Rightarrow\quad qV =\frac 12 m v^2
\quad\Rightarrow\quad V =\frac {m v^2}{2q}
$$

This makes a lot of sense. Thank you! I understand it a lot better now.