Oh jesus. potential difference problems help per favore

[smd1991]

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

 

[Astronuc]

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.

 

[Moetasim]

1) To calculate the work done, we can use the equation W=qEd, where q is the charge in coulombs, E is the electric field, and d is the distance. In this case, we have a potential difference of 6 volts and a charge of 0.19 coulombs. Therefore, the work done would be:

W= (0.19 coulombs)(6 volts) = 1.14 Jules

2) To calculate the potential difference needed for an electron to achieve a speed of 6% of the speed of light, we can use the equation V=PE/q, where PE is the potential energy and q is the charge of the electron. The potential energy can be calculated using the equation PE=qEd, where E is the electric field and d is the distance. In this case, the electron is starting from rest, so the potential energy is equal to the kinetic energy at the given speed (since PE at rest is 0). Therefore, the potential energy would be:

PE= (1/2)(9.11x10^-31 kg)(0.06c)^2 = 1.305x10^-19 Jules

Now, we can plug this value into the equation V=PE/q, where q is the charge of an electron (1.602x10^-19 coulombs):

V= (1.305x10^-19 J)/(1.602x10^-19 coulombs) = 0.815 volts

Therefore, the potential difference needed for an electron to achieve a speed of 6% of the speed of light would be 0.815 volts.