Electric Potential and Energy Storage

Question

The plates of a certain parallel-plate capacitor have been set with a separation of d = 4.15 mm, and each has an area of A = 50 cm². After charging the capacitor, each plate carries a full charge of magnitude Q = 1.55 x 10^-7 C. The capacitor is then disconnected from the power supply.

If the plates are in vacuum, what is the potential difference between the plates? Please explain your process.

What is the energy stored between the plates? Please, explain your process.

Answer 1

Hello Kat,

To find the potential difference between the plates of the parallel-plate capacitor and the energy stored between them, you can use the following formulas:

1. Potential Difference (V) between the plates:

V = Q / C

2. Energy (U) stored between the plates:

U = (1/2) * C * V^2

Let's calculate them step by step:

1. Potential Difference (V):

The potential difference between the plates of the capacitor is equal to the charge (Q) on one of the plates divided by the capacitance (C). The capacitance of a parallel-plate capacitor is given by the formula C = ε₀ * (A / d), where ε₀ is the vacuum permittivity, A is the area of the plates, and d is the separation distance between the plates.

ε₀ (vacuum permittivity) is approximately 8.85 x 10^(-12) F/m.

Calculate C:

C = ε₀ * (A / d)

C = (8.85 x 10^(-12) F/m) * (0.005 m^2 / 0.00415 m)

C = 0.01064 F (Farads)

Now, use the potential difference formula to find V:

V = Q / C

V = (1.55 x 10^(-7) C) / 0.01064 F

V ≈ 1.46 x 10^4 V (volts)

2. Energy Stored (U):

The energy stored between the plates of the capacitor can be calculated using the formula mentioned earlier:

U = (1/2) * C * V^2

Substitute the values:

U = (1/2) * 0.01064 F * (1.46 x 10^4 V)^2

U ≈ 1.43 x 10^(-4) joules

So, the potential difference between the plates is approximately 1.46 x 10^4 V (volts), and the energy stored between the plates is approximately 1.43 x 10^(-4) joules.

Hope this helps out, feel free to send me a message if you have one