Question

Two large parallel conducting plates of area 0.24 m2 are separated by a distance of 0.5 mm. The plates carry opposite charges, and the electric field in the space between them is 4.70 105 V/m. What is the electric energy? J

Answer 1

Answer 2

Electric Energy Between Parallel Plates

Given:

• Plate area ($A$) = $0.24{\text{m}}^2$

• Separation distance ($d$) = $0.5\text{mm}=0.5\times 10^{-3}\text{m}$

• Electric field ($E$) = $4.70\times 10^5\text{V/m}$

Formulas:

1. Energy density ($u$) in an electric field:

   $$u=\frac{1}{2}{ϵ}_0{E}^2$$

   (${ϵ}_0=8.85\times 10^{-12}\text{F/m}$)

2. Total energy ($U$) stored:

   $$U=u\times \text{Volume}=\frac{1}{2}{ϵ}_0{E}^2\times (A\times d)$$

Calculation:

1. Energy density:

   $$u=\frac{1}{2}(8.85\times 10^{-12})(4.70\times 10^5{)}^2=0.5\times 8.85\times 10^{-12}\times 2.209\times 10^{11}$$$$u=0.5\times 1.955=0.9775{\text{J/m}}^3$$

2. Total energy:

   $$U=0.9775\times (0.24\times 0.5\times 10^{-3})$$$$U=0.9775\times 1.2\times 10^{-4}=1.173\times 10^{-4}\text{J}$$

Answer:
The electric energy stored is $1.17\times 10^{-4}\text{J}$ (or 117 μJ).