Voltage that brings electron to rest

[Alexander2357]

Homework Statement

An electron is moving at $$5.4 \times 10^5 ms^(-1)$$ is brought to rest as it enters a region of lower potential. What is the voltage that brought it to rest?

Homework Equations

$$\Delta KE=1/2m(v_2)^{2} - 1/2m(v_1)^{2} = W = -\Delta U = q \times \Delta V$$

The Attempt at a Solution

I substituted numbers into the equation and got $$\Delta KE = -1.315 \times 10^{-19} J$$, then I divided this by $$-1.602 \times 10^{-19} C$$ and got $$0.83 V$$ but the answer is wrong according the the answers in the back of the book. Why is this?

 

[TSny]

Hello, Alexander2357. Looks like you have the right idea. But..

$$\Delta KE=1/2m(v_2)^{2} - 1/2m(v_1)^{2} = W = -\Delta U = \frac{q}{\Delta V}$$

Is the very last expression of this equation correct? Looks like maybe a careless slip. You should also consider if you have the right sign for this expression.

Also, did you state the question exactly as it was given to you? The question "What is the voltage that brought it to rest?" is a little ambiguous to me. I think it would have been better to phrase it "What is the potential difference that brought it to rest?". I'm just wondering if you are expected to consider the sign of the potential difference that the electron moves through.

 

[D.R.U]

What is answer of book?

 

[Alexander2357]

Hello, Alexander2357. Looks like you have the right idea. Is the very last expression of this equation correct?

I messed up the LaTeX. Fixed it.

I used the correct expression in my calculations anyway.

 

[TSny]

$$\Delta KE=1/2m(v_2)^{2} - 1/2m(v_1)^{2} = W = -\Delta U = q \times \Delta V$$

Is the sign of the last expression correct?

 

[D.R.U]

I got answer 0.82V...what is actual answer?

 

[ehild]

The problem is with the sign of the potential (with respect to the potential at the initial position of the electron). Should it be minus or plus to stop the electron?

ehild

 

[Alexander2357]

The problem is with the sign of the potential (with respect to the potential at the initial position of the electron). Should it be minus or plus to stop the electron?

ehild

I am not sure. May be I am not understanding the concept of a potential properly.

Why should the potential be negative to stop the electron?

This isn't a textbook question, it is a quiz question. I don't know why I said it is a textbook question... so I don't know the right answer but its magnitude should be ~0.826.

I entered 0.83 (2 sig fig) and it wasn't correct.

 

[ehild]

If a plate charged to positive is opposite to the electron, will it attract or repulse the electron?

ehild

 

[Alexander2357]

If a plate charged to positive is opposite to the electron, will it attract or repulse the electron?

ehild

I forgot that potential has the same sign as the charge that produces it. So the potential must be negative to repulse the electron and slow it down. I think it makes sense now.

 

[ehild]

OK. Is -0.83 V accepted?

 

[Alexander2357]

Yes! Thank you very much everyone!