# Two protons are released from rest when they are 0.700 nm apart. What is the maximum speed they...

## Question:

Two protons are released from rest when they are 0.700 nm apart. What is the maximum speed they will reach?

## Electric Potential Energy:

In general we can define potential energy as stored energy that is available for later use. In the case of at least two point charges, we can determine an electrical potential energy of the form:

{eq}\displaystyle U_1 = \frac{kq_1 q_2}{r} {/eq}

where:

- {eq}\displaystyle k = 9\ \times\ 10^9\ Nm^2/C^2 {/eq} is the Coulomb constant

- {eq}\displaystyle q_1 {/eq} and {eq}\displaystyle q_2 {/eq} are the two charges

- {eq}\displaystyle r {/eq} is the distance between the two charges

## Answer and Explanation:

Given:

- {eq}\displaystyle r = 0.7\ nm = 7\ \times\ 10^{-10}\ m {/eq} is the distance between the protons

We can use the conservation of energy to solve this problem. So here in general we can write the kinetic energy *K* and potential energy *U* as:

{eq}\displaystyle K_1 + U_1 = K_2 + U_2 {/eq}

When the protons have not yet moved before they are released, all of the energy is potential. And since the electric potential energy varies inversely with *r*, this means that when *r* is at infinity, the total energy is kinetic. We thus have:

{eq}\displaystyle U_1 = K_2 {/eq}

We expand these terms as:

{eq}\displaystyle \frac{kq_1 q_2}{r} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 {/eq}

We have two kinetic energy terms here since we have two protons. Note that both protons will move due to each other's repulsive forces, so both of them have to divide the total energy between themselves. So here we can simplify this as:

{eq}\displaystyle mv^2 = \frac{kq_1 q_2}{r} {/eq}

We isolate the speed *v*:

{eq}\displaystyle v = \sqrt{\frac{kq_1 q_2}{mr}} {/eq}

Now for a proton, we have:

- {eq}\displaystyle m = 1.67\ \times\ 10^{-27}\ kg {/eq}

- {eq}\displaystyle q = 1.602\ \times\ 10^{-19}\ C {/eq}

We substitute:

{eq}\displaystyle v = \sqrt{\frac{(9\ \times\ 10^9\ Nm^2/C^2)(1.602\ \times\ 10^{-19}\ C)^2}{(1.67\ \times\ 10^{-27}\ kg)(7\ \times\ 10^{-10}\ m)}} {/eq}

We thus get:

{eq}\displaystyle \boxed{v = 14,056.48\ m/s} {/eq}

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from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6