A 24.0-V battery is connected to a 4.70-µF capacitor. How much energy (in J) is stored in the capacitor?
I'm guessing this is really easy but my professor sent us into the homework blind...again...
C'mon. In these days, you are never blind. A simple web search immediately points to
http://farside.ph.utexas.edu/teaching/302l/lectures/node47.html
where the whole topic is discussed, and formulas are derived.
Surely you have a text for the course, where this is treated. Might try reading it.
I do have the text but it is by far the most worthless physics textbook available at the college level, no one in the class uses it, it just makes the class more confusing, though he does let us use it on the quizzes/tests (which sometimes helps). Thank you for the link it really helped, I tried doing an internet search but I'm on a campus computer right now (with a LOT of blocked content) so not much came up.
To calculate the energy stored in a capacitor, you can use the formula:
E = (1/2) * C * V^2
Where:
E = Energy stored in the capacitor (in joules)
C = Capacitance (in farads)
V = Voltage across the capacitor (in volts)
In this case, the voltage across the capacitor is 24.0 V, and the capacitance is 4.70 µF (microfarads). However, it's important to note that the formula requires the capacitance to be in farads.
Since 1 µF = 10^(-6) F (farads), we need to convert the capacitance to farads:
C = 4.70 µF = 4.70 * 10^(-6) F
Now we can calculate the energy stored in the capacitor:
E = (1/2) * C * V^2
= (1/2) * (4.70 * 10^(-6) F) * (24.0 V)^2
Let's plug in the values and calculate the answer.