# How Do You Calculate the Potential Difference to Accelerate a He+ Ion?

• BuBbLeS01
In summary, to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 2690000 m/s, a potential difference of 2.4x10^-14 V is needed.
BuBbLeS01

## Homework Statement

What potential difference is needed to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 2690000 m/s?

## The Attempt at a Solution

Change in V = Change in K
Change in V = 1/2 mv^2
But what dose mass = 4u mean? What is u?

Oh ok...
so is the rest of the problem set up correctly?
So I do...
1/2 * (4 * 1.66x10^-27) * (2690000^2) = 2.4x10^-14 V

BuBbLeS01 said:
Oh ok...
so is the rest of the problem set up correctly?
So I do...
1/2 * (4 * 1.66x10^-27) * (2690000^2) = 2.4x10^-14 V

actually.. i think you have something fundamentally wrong with:

change in V = change in K

this is because, the equation is dimensionally incorrect. 'K' or it's change, has the units of energy, whereas 'V' has the units of 'Energy/Charge'.

your solution maybe right.. or i don't know what.. just explain to me why you took the above mentioned relation.

You need to think of this problem in terms of the Electric field? For a given potential difference, what is the field present? Or to think in the direction of the problem, what is the electric field required so that the necessary velocity is achieved?

I think maybe it should be...
Change in K = q*Change in V

BuBbLeS01 said:
I think maybe it should be...
Change in K = q*Change in V

and yes.. you're right. Now, just substitute the values of 'm', 'v', 'q' and find what 'V' comes out to be.

also.. don't use the terminology 'change in V'. 'V' in itself refers to the potential difference.

Change in 'K', basically means the total Work done [following by 'Work-Energy theorem']. And 'V' is defined as the amount of work done per unit charge. So what you effectively did was first get the amount of work required to accelerate the given particle and then equate it with the amount of potential that could provide such energy [or in other words, do this work].

Change in K = q*Change in V
(1/2 (6.6423x10^-24) * 26900000^2) / (1.6x10^-19) = V = 1.5x10^10 V
Is that right then?

Oh wait to many 0's is velocity...answer I got is 1.5x10^8 V

OMG...wrong number for mass lol...

(1/2 (6.6423x10^-27) * 2690000^2) / (1.6x10^-19) = V = 1.5x10^5 V
Is that right then?

oh nooo its wrong :( I only have 1 try left...what did I do wrong?

BuBbLeS01 said:
OMG...wrong number for mass lol...

(1/2 (6.6423x10^-27) * 2690000^2) / (1.6x10^-19) = V = 1.5x10^5 V
Is that right then?

well.. seems right to me..

also.. try not to double post. Use the  button in the bottom right corner to edit ur latest reply in case u have to add something. Make a new reply only when somebody has already commented on something or ur replies are far apart. Your last 3 replies were less than 5 mins. apart. Just keeps the forums clean. No offense.

EDIT:

i can't see anything wrong with the above. Are you sure it's a $He^+$ ion and not a $He^{+2}$ nucleus?

Last edited:

## What is electric potential difference?

Electric potential difference, also known as voltage, is the difference in electric potential between two points in an electric field. It is measured in volts (V) and represents the amount of energy required to move a unit of electric charge from one point to another.

## How is electric potential difference measured?

Electric potential difference is measured using a voltmeter, which is connected in parallel to the two points between which the potential difference is being measured. The voltmeter measures the difference in electric potential between the two points and displays it in volts.

## What is the relationship between electric potential difference and electric current?

Electric potential difference is directly proportional to electric current. This means that as the potential difference increases, the current also increases. This relationship is described by Ohm's Law, which states that the current flowing through a conductor is directly proportional to the potential difference across it.

## What factors affect electric potential difference?

The main factors that affect electric potential difference are the distance between two points and the amount of charge present. The greater the distance between two points, the higher the potential difference. Similarly, the greater the amount of charge present, the higher the potential difference.

## Why is electric potential difference important?

Electric potential difference is important because it is necessary for the flow of electric current. Without a potential difference, there would be no movement of electric charges and therefore no electricity. It also plays a crucial role in various electrical applications, such as powering electronic devices and distributing electricity in power grids.

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