Final answer:
The voltage across the capacitor is 1600V, and the capacitance is 9.375 μF, derived from the equations of energy stored in a capacitor in terms of charge and voltage.
Step-by-step explanation:
To calculate the voltage and capacitance of the capacitor using the given values of charge and energy stored, we can use the relationship between these quantities. The energy (U) stored in a capacitor is given by the equation U = ½ CV², where C is the capacitance in farads, V is the voltage in volts, and U is the energy in joules.
First, rearrange the formula to solve for V:
V = √(2U/C). Substituting the known values (U = 12 J and charge q = 15 mC = 15 × 10⁻³ C) into the equation, we have to find C using the relation between charge and capacitance which is C = q/V. Since we don't have V directly, let's first express the energy in terms of charge: U = ½ qV, which can be rearranged to find V: V = (2U)/q.
Then, we find the voltage as V = (2·12J)/(15×10⁻³C) = 1600V.
Next, we can calculate the capacitance by rearranging q = CV to find C = q/V which gives C = (15×10⁻³ C)/(1600 V) = 9.375 × 10⁻¶ F or 9.375 μF.