Two parallel circular plates, each with a radius of $7.0 \mathrm {~mm}$ and carrying equal-magnitude surface charge densities of $\pm 3.0 \mu \mathrm { C } / \mathrm { m } ^ { 2 }$ are separated by a distance of $1.0 \mathrm {~mm}$ with only air between them. How much energy is stored in these plates? (?₀ = 8.85 × 10^-12C²/N ? m²)

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by:$$U = \frac{1}{2}CV^2$$where C is the capacitance and V is the voltage across the capacitor.The capacitance of a parallel-plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where $\epsilon_0$ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.In this problem, we have:$$\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2$$$$A = \pi (7.0 \times 10^{-3} \text{ m})^2 = 1.54 \times 10^{-4} \text{ m}^2$$$$d = 1.0 \times 10^{-3} \text{ m}$$$$\sigma = \pm 3.0 \mu \text{C}/\text{m}^2 = \pm 3.0 \times 10^{-6} \text{ C}/\text{m}^2$$The voltage across the capacitor is given by:$$V = \frac{\sigma d}{\epsilon_0}$$Substituting the given values into the equations above, we get:$$C = \frac{(8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2)(1.54 \times 10^{-4} \text{ m}^2)}{1.0 \times 10^{-3} \text{ m}} = 1.36 \times 10^{-11} \text{ F}$$$$V = \frac{(3.0 \times 10^{-6} \text{ C}/\text{m}^2)(1.0 \times 10^{-3} \text{ m})}{8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2} = 342 \text{ V}$$Substituting the values of C and V into the equation for energy, we get:$$U = \frac{1}{2}(1.36 \times 10^{-11} \text{ F})(342 \text{ V})^2 = 78 \text{ nJ}$$Therefore, the energy stored in the plates is 78 nJ.

Two Parallel Circular Plates, Each with a Radius Of 7.0 mm7.0 \mathrm {~mm}7.0 mm

Two Parallel Circular Plates, Each with a Radius Of $7.0 \mathrm {~mm}$