When is the Charge and Electric Energy in a Capacitor Half of Its Final Value?

[slaytanic37]

Homework Statement

part 1. Consider a simple RC circuit consisting of a resistor and a capacitor connected in series to a battery. The resistor has a value of 8kOhms, and the capacitance is 3uF. The final (asymptotic) value of the charge in the capacitor is E*C. At what time is the charge in the capacitor one half of this final value?

part 2. At what time is the electric energy in the capacitor one-half of its final value?

Homework Equations

Q(t) = EC(1-e^(t/(RC)))
E = Voltage
C = Capacitance
t = time

The Attempt at a Solution

for part 1, i set (1-e^(t/(RC))) = .5*EC and got a time of .0218.
for part 2, i don't know the equation to relate electric energy of a capacitor over time. i think that's all i need is just the equation.

thanks in advance!

 

[LowlyPion]

Welcome to PF.

Maybe this will give you an idea or three?
http://en.wikipedia.org/wiki/Capacitor#Energy_storage

 

[nlightner]

I would like to clarify that the equations used in the attempt at a solution are correct and applicable for the given circuit. However, the time calculated for part 1 may not be entirely accurate due to rounding errors. It would be helpful to provide the units for the time in the solution.

For part 2, the electric energy in a capacitor can be calculated using the equation E = 1/2 * CV^2, where V is the voltage across the capacitor. As the voltage across the capacitor is directly proportional to the charge in the capacitor, the time at which the electric energy is one-half of its final value would be the same as the time at which the charge in the capacitor is one-half of its final value. Therefore, the time calculated for part 1 can be used for part 2 as well.

In conclusion, the equations and approach used in the attempt at a solution are accurate, but it is important to provide units for the calculated time and to note the relationship between charge and voltage when calculating the electric energy in a capacitor.