Final answer:
The capacitance of the capacitor is 22.4 µF (microfarads), calculated by rearranging the energy stored in a capacitor formula and inserting the given charge and energy values.
Step-by-step explanation:
The question is asking for the capacitance of a capacitor that stores a charge of 5.36e-5 coulombs and an energy of 6.45e-5 joules. To find the capacitance (C), we can use the formula for the energy (E) stored in a capacitor: E = (1/2) * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. However, we do not have the voltage directly, so we need to use the formula that relates charge (Q) to capacitance and voltage: Q = C * V. Combining the two equations, we can solve for capacitance with the given charge and energy values.
Finding the Capacitance
First, rewrite the energy formula in terms of Q:
E = Q^2 / (2 * C).
Then, isolate C and solve:
C = Q^2 / (2 * E).
Putting the given values into the formula, we get:
C = (5.36e-5 C)^2 / (2 * 6.45e-5 J) = 2.24e-5 F or 22.4 µF.