# Electrical Potential Force

• stylez03
In summary: You need meters per second to get Newtons.In summary, we used the given information of two protons with speeds of 1350 km/s aimed directly towards each other by a cyclotron accelerator to find the maximum electrical force that these protons will exert on each other. We first found the distance of closest approach by setting the initial kinetic energy (total kinetic energy of both protons) equal to the electric potential energy, using the magnitude of the potential energy formula. We then used Coulomb's Law, with R representing the distance of closest approach, to find the force exerted on the protons. However, in order to use Coulomb's Law, we needed to convert the given speeds from km/s to m/s. Once

## Homework Statement

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1350 km/s, measured relative to the earth.

Find the maximum electrical force that these protons will exert on each other.

## Homework Equations

Mass of Proton $$1.67 * 10 ^{-27}$$

$$KE_{i} = PE_{f}$$

$$\frac {1} {2} * m_{p}*v_{1}^{2} = \frac {1} {4*pi*8.85*10^{-12}}*\frac {Q_{2}} {R}$$

## The Attempt at a Solution

I'm not sure where to go from here.

Find the distance of closest approach by setting the initial KE (total KE) equal to electric potential energy. Then use Coulomb's law to find the force.

Doc Al said:
Find the distance of closest approach by setting the initial KE (total KE) equal to electric potential energy. Then use Coulomb's law to find the force.

What is V_i ?

stylez03 said:
What is V_i ?
What do you mean by V_i? The initial speed? That's given. The initial potential? Assume they start out infinitely far away from each other (for all practical purposes).

Doc Al said:
What do you mean by V_i? The initial speed? That's given. The initial potential? Assume they start out infinitely far away from each other (for all practical purposes).

To find the closet approach, I'm solving for R as you said so it should be in the form of:

$$\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}} = R$$

Looks to me like you are only counting the KE of one proton--use the total KE of both.

Doc Al said:
Looks to me like you are only counting the KE of one proton--use the total KE of both.

When we count both, would it just be KE_1 + KE_2?

$$2*(\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}}) = R$$

That's right--just add up the KE. But fix that equation--I just realized that you have some factors inverted. The magnitude of the PE is kq^2/R.

Doc Al said:
That's right--just add up the KE. But fix that equation--I just realized that you have some factors inverted. The magnitude of the PE is kq^2/R.

Ah let's show each step:

$$KE_{i}1 + KE_{i}2 = PE_{f}$$

$$2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R}$$

$$2(m_{p}*v_{i}) = \frac {KQ^{2}} {R}$$

We divide out KQ^2 to isolate R

$$\frac {2(m_{p}*v_{i})} {KQ^{2}} = {R}$$

??

stylez03 said:
$$KE_{i}1 + KE_{i}2 = PE_{f}$$
Good.

$$2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R}$$
Typos in the right hand term.

$$2(m_{p}*v_{i}) = \frac {KQ^{2}} {R}$$
Almost--that should be v_i^2. And the 2 cancels.

We divide out KQ^2 to isolate R

$$\frac {2(m_{p}*v_{i})} {KQ^{2}} = {R}$$
Do this step over. Note that R starts out in the denominator.

Last edited:
$$m_{p}*v_{i} = \frac {KQ^{2}} {R}$$

Divide by $$m_{p}*v_{i}$$

$$\frac {KQ^{2}} {(m_{p}*v_{i})*R}$$

Multiply by R

$$R = \frac {KQ^{2}} {(m_{p}*v_{i})}$$

Sorry my algebra is a little rusty

stylez03 said:
$$m_{p}*v_{i} = \frac {KQ^{2}} {R}$$
Good.

Divide by $$m_{p}*v_{i}$$

$$\frac {KQ^{2}} {(m_{p}*v_{i})*R}$$
You mean:

$$1 = \frac {KQ^{2}} {(m_{p}*v_{i})*R}$$

Multiply by R

$$R = \frac {KQ^{2}} {(m_{p}*v_{i})}$$
Good.

Sorry my algebra is a little rusty
No problem. Remember the cross-multiply trick. When you see this:

$$\frac{A}{B} = \frac{C}{D}$$

You can convert to this:

$$A*D = B*C$$

After I find the closet approach, you said to use Coulomb's Law to find the force. Is the R in here equilv to d in Coulomb's Law where:

$$F = \frac {K*Q_{1}*Q_{2}} {d^{2}}$$

and F is what I'm after in t his problem?

Yes and yes.

$$R = \frac {KQ^{2}} {(m_{p}*v_{i}^{2})}$$

$$R = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(1.67*10^{-27})*(1350)^{2}}$$

$$R = \frac {2.30*10^{-28}} {4.50*10^{-24}}$$

$$R = 0.000051$$

$$F = \frac {K*Q_{1}*Q_{2}} {d^{2}}$$

$$F = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(0.000051)^{2}}$$

$$F = 8.84 * 10^{-20}$$

Can someone verify my calculations, the online program says it's incorrect. I think I did everything right, I'm wondering if km/s needs to be converted into m/s?

Last edited:
stylez03 said:
... I'm wondering if km/s needs to be converted into m/s?
Absolutely.

## 1. What is electrical potential force?

Electrical potential force, also known as electric potential energy or electrostatic potential energy, is the potential energy that an object possesses due to its position in an electric field.

## 2. How is electrical potential force calculated?

The electrical potential force is calculated by multiplying the electric charge of an object by the electric potential difference, or voltage, between two points in the electric field.

## 3. What are the units of electrical potential force?

The SI unit for electrical potential force is joules per coulomb (J/C), but it is often expressed in volts (V) for convenience.

## 4. What is the difference between electrical potential force and electric field?

Electrical potential force is a measure of the potential energy an object has in an electric field, while electric field is a measure of the force that a charged particle would experience in that field.

## 5. How does distance affect electrical potential force?

The electrical potential force between two charged objects decreases as the distance between them increases. This is because the electric field weakens with distance, resulting in a decrease in the potential energy between the two objects.