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# Two problems that I'm stumped by. How much, and how heavy?

Marcus recharges the dead 12.0 volt battery by sending 2.8 x 10^4 C of charge through the terminals. How much potential energy must Marcus store in the car battery to make it fully charged?

An oil drop having a charge of 8.0 x 10^-19 C is suspended between two charged parallel plates. The plates are separated by a distance of 8.0 mm, and there is a potential difference of 1200 V between the plates. What is the weight of the suspended oil drop?

Assistance would be hot. These are the only two questions for this lesson I haven't been able to figure out.

### 1 Answer

- oldprofLv 71 decade agoFavorite Answer
The charge of a single electron q = 1.6 * 10-19 C. So the fully charged battery will have PE = (Q/q)V = (2.8E4/1.6E-19)*12 = 2.1E24 eV or 336460 Joules or 94 Watt-hrs. Q = 2.8E4 coulombs of charge when fully charged. Q/q = n the number of electrons; so nV = electron volts which are units of energy.

The E field between two parallel plates is E = V/h, where V = 1200 V and h = 8E-6 m. When suspended the weight w = mg = qE = qV/h = f the force pulling on the charge q = 8E-19 C. Solve for w = 8E-19*1200/8E-6 = E-13*1200 = 12E-11 Newtons.