Oh really? potential difference sucks

AI Thread Summary
To calculate the work done in transferring 0.19 coulombs of charge through a potential difference of 6 volts, use the formula W = qV, resulting in 1.14 joules. For the second question, to find the potential difference needed to accelerate an electron to 6% of the speed of light, one must consider the kinetic energy formula and the relationship between kinetic energy and potential difference. The required potential difference can be derived from the electron's mass and the desired speed. Both calculations rely on fundamental principles of electromagnetism and energy conservation. Understanding these concepts is essential for solving problems related to electric potential and charge movement.
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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