Can someone please help me with this, I want it similar to the example I have posted down below. I need help with the solution and summary and application part please

Question:

The electric field between two parallel plates is 1500 N/C. If the plates are 2.5 mm apart and have a capacitance of 40 F, how much charge do the plates store?

Answer: 1.5 x10^-4

Solution:

v=Ed

v= 1500 x 2.5E-3 = 3.75

Q= C x V

= 40 x 10^-6 x 3.75

= 1.5 x 10^-4 C

Example

Question: A circuit consists of a battery with an EMFof 12 V and three capacitors in series, where C1 = 15 ,uF, C2 =30 ,uF, and C3 = 45 ,uF. What energy is stored in the capacitors? Summary: The main content required for this question is the equation for the potential energy of a capacitor once it is charged. This equation is U = (1/2JCV'2. All the information that we need to calculate this problem is the capacitance and the EM F of the capacitor. Solution: First, we must solve for the capacitance, by summing the individual capacitance values given. Since the capacitators are in a series, we do this by summing them such as: C(tot) = [(1/C1) + (1/C2) + (1/C3}]"-1 = [(1/15E-6) + (1/30E-6) + (1/45E-6)]"-1 = 8.18E-6 F Plug this value into our equation. given that our EMF is 12 V; U = (1/2) (8.18E-6 F) (12 VJ"2 Our answer is; = 5.9E-4J Application: A common application of energy stored within capacitors is with AED machines, or more commonly known as defibrillators. The energy stored within the defibrillator that is used to electrically restart someone's heart, is stored within a capacitor! We use a small monitor to determine how much energy to transfer out of the capacitor and into the defibrillator itself, and that potential energy that was stored in the capacitor is delivered into someone's body! Sources: Energy Stored in Capacitors. (2022, February 20). https://phys.libretexts.org/@go/page/2676