We read your photo as:

Calculate the energy a parallel plate capacitor (charged to a voltage of V=12V) with plate area A = 2m2, plate separation d = 0.02m, and vacuum permittivity ε₀ = 8.854 × 10⁻¹² F/m.

Physics

ScanSolve AI Answer

Step-by-step solution

2 steps

Here are the solutions to the questions:

QUESTION ONE

b) Calculate the energy a parallel plate capacitor (charged to a voltage of V=12VV=12\text{V}) with the following:

  • Plate area A=2m2A = 2\text{m}^2
  • Plate separation d=0.02md = 0.02\text{m}
  • Vacuum permittivity ϵ0=8.854×10−12 F/m\epsilon_0 = 8.854 \times 10^{-12}\text{ F/m}
  1. Calculate the capacitance CC of the parallel plate capacitor.

    The formula for the capacitance of a parallel plate capacitor in vacuum is:

    C=ϵ0AdC = \frac{\epsilon_0 A}{d}

    Substitute the given values:

    C=(8.854×10−12 F/m)×(2 m2)0.02 mC = \frac{(8.854 \times 10^{-12}\text{ F/m}) \times (2\text{ m}^2)}{0.02\text{ m}} C=17.708×10−120.02 FC = \frac{17.708 \times 10^{-12}}{0.02}\text{ F} C=885.4×10−12 FC = 885.4 \times 10^{-12}\text{ F} C=885.4 pFC = 885.4\text{ pF}
  2. Calculate the energy WeW_e stored in the capacitor.

    The formula for the energy stored in a capacitor is:

    We=12CV2W_e = \frac{1}{2} C V^2

    Substitute the calculated capacitance CC and the given voltage VV:

    We=12×(885.4×10−12 F)×(12 V)2W_e = \frac{1}{2} \times (885.4 \times 10^{-12}\text{ F}) \times (12\text{ V})^2 We=12×(885.4×10−12)×144 JW_e = \frac{1}{2} \times (885.4 \times 10^{-12}) \times 144\text{ J} We=442.7×10−12×144 JW_e = 442.7 \times 10^{-12} \times 144\text{ J} We=63748.8×10−12 JW_e = 63748.8 \times 10^{-12}\text{ J}
Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp