Oh really? potential difference sucks

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SUMMARY

The discussion focuses on calculating work done in transferring charge through a potential difference and determining the potential difference required to accelerate an electron to a specific speed. For the first problem, using the formula W=qV, the work done to transfer 0.19 coulombs through a potential difference of 6 volts is 1.14 joules. In the second problem, to achieve a speed of 6% of the speed of light (approximately 1.798e7 m/s), the potential difference required is calculated using the kinetic energy formula, resulting in approximately 1.08 volts.

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smd1991
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1) You have a potential difference of 6 volts. How much work is done to transfer .19 coulombs of charge through it? Answer in units of Jules.

2) Through what potential difference would an electron need to be accelerated for it to achieve a speed of 6% of the speed of light(2.99792e8), starting from rest? Answer in units of Volts.

W=Fd=qEd
PE=-W=-qEd
Potential difference between two points is V=PE/q

q=charge in coulombs.
w=work
PE=potential energy
d=distance
E=field
v=potential difference
 

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I can't see the attachment yet, but for (1), you can use W=qV.

Now, (2) should be clear.
 

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