Oh jesus. potential difference problems help per favore

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 problem, to find the potential difference required for an electron to reach 6% of the speed of light, one must first calculate the kinetic energy using the classical formula KE = 0.5mv², where m is the mass of the electron. The potential difference can then be determined by equating the kinetic energy to the work done by the electric field, which is given by the formula V = KE/q. The discussion emphasizes the relationship between potential difference, work, and energy in electric fields.
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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Please show some work on the problem.

In the first problem, 1 Volt (of potential difference) applies 1 Joule/coulomb of work to a charge.

In the second problem, knowing the velocity of the charge, one may compute the kinetic energy (here one may need to consider relativistic effects), or one can make a rough approximation with the definition of kinetic energy from classical mechanics.

The increase in kinetic energy is equal to the work done by the electric field.
 
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