A Proton is Accelerated Through What Voltage?

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The discussion revolves around calculating the voltage required to accelerate a proton to a speed of 8.2 x 10^4 m/s towards a charged metal sphere of 30 cm diameter with a charge of +1.00 nC. The calculations involve using Coulomb's law (K0 = 8.99 x 10^9 N m²/C²) and the kinetic energy formula. The initial voltage calculated was 59.93V, and the change in voltage (Delta V) was computed as 35.09V, leading to a total of 95V. However, the user expresses confusion over the discrepancy in the expected answer, indicating a misunderstanding of the kinetic energy principle.

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chunkytuna21
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A metal target sphere of 30 cm diameter suspended out in space is given a charge of +1.00 nC. Through what voltage must a distant proton be accelerated from rest (by some sort of space weapon) if it is to arrive at the sphere at a speed of 8.2*10^4 m/s? (Neglect any gravitational effects.)

K0= 8.99*10^9, mass of proton= 1.67*10^-27kg, qe= 1.6*10^-19
V= (K0*Q)/R
1/2mvf^2= qe*delta V

First I calculated V: (8.99*10^9)(1.00*10^-9)/0.15m= 59.93V
Then I calculated Delta V: 1/2(1.67*10^-27)(82000^2)/1.6*10^-19= 35.09V
59.93V+35.09= 95V

Now, this is where I am confused. The answer is not 35V or 59V or even 95V. I really can't see any other way to solve this problem. Where am I going wrong? Any help is greatly appreciated.
 
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chunkytuna21 said:
Where am I going wrong?
Probably right here:
chunkytuna21 said:
1/2mvf^2= qe*delta V
Write the general principle that you're using here.
 
What do you mean? That is the equation for kinetic energy of the proton. I don't know the principle behind it.
 

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