Question

Calculate the largest possible electrostatic energy in 1 cm³ volume of air. The dielectric breakdown of air happens when the field exceeds 3 × 10⁶ V/m.

• A. 39.6 J
• B. 39.6 μJ
• C. 3.96 J
• D. 3.96 μJ

Answer 1

Answer: (B)

39.6 μJ

Answer 2

The problem asks us to calculate the largest possible electrostatic energy stored in a 1 cm³ volume of air before dielectric breakdown occurs.

1. Identify Given Parameters:

• Volume of air, $V = 1 \, \text{cm}^3 = 1 \times (10^{-2} \, \text{m})^3 = 10^{-6} \, \text{m}^3$.
• Dielectric breakdown field of air, $E_{max} = 3 \times 10^6 \, \text{V/m}$.

2. Identify Relevant Physical Constants:

• The permittivity of air can be approximated by the permittivity of free space, $\epsilon_0$. $\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}$. To match the given options precisely, we can use the approximation $\epsilon_0 = 8.8 \times 10^{-12} \, \text{F/m}$, which is sometimes used in problems.

3. Formula for Electrostatic Energy Density:

The electrostatic energy density ($u$) in a dielectric medium is given by: $u = \frac{1}{2} \epsilon E^2$ where $\epsilon$ is the permittivity of the medium and $E$ is the electric field strength.

4. Calculate Maximum Energy Density:

The largest possible electrostatic energy occurs when the electric field reaches its maximum possible value, which is the dielectric breakdown field $E_{max}$. So, the maximum energy density $u_{max}$ is: $u_{max} = \frac{1}{2} \epsilon_0 E_{max}^2$

Substitute the values: $u_{max} = \frac{1}{2} \times (8.8 \times 10^{-12} \, \text{F/m}) \times (3 \times 10^6 \, \text{V/m})^2$ $u_{max} = \frac{1}{2} \times 8.8 \times 10^{-12} \times (9 \times 10^{12}) \, \text{J/m}^3$ $u_{max} = 4.4 \times 9 \times 10^{(-12+12)} \, \text{J/m}^3$ $u_{max} = 39.6 \times 10^0 \, \text{J/m}^3$ $u_{max} = 39.6 \, \text{J/m}^3$

5. Calculate Total Electrostatic Energy:

The total electrostatic energy $U_{max}$ in the given volume $V$ is the product of the maximum energy density and the volume: $U_{max} = u_{max} \times V$ $U_{max} = (39.6 \, \text{J/m}^3) \times (10^{-6} \, \text{m}^3)$ $U_{max} = 39.6 \times 10^{-6} \, \text{J}$ $U_{max} = 39.6 \, \mu\text{J}$

Comparing this result with the given options, option B matches our calculated value.