A 3.0-pFcapacitor consists of two large closely-spaced parallel plates that have surface charge densities of $\pm 1.0 \mathrm { nC } / \mathrm { mm } ^ { 2 }$ If the potential across the plates is $23.0 \mathrm { kV }$ with only air between them, find the surface area of each of the plates. (?₀ = 8.85 × 10^-12C²/N ? m²)

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in farads (F)* $\epsilon_0$ is the permittivity of free space (8.85 × 10^-12 C²/N m²)* A is the surface area of each plate in square meters (m²)* d is the distance between the plates in meters (m)We are given that C = 3.0 pF = 3.0 × 10^-12 F, $\sigma$ = $\pm 1.0 \mathrm { nC } / \mathrm { mm } ^ { 2 }$ = $\pm 1.0 \times 10^{-9} \mathrm { C } / \mathrm { m } ^ { 2 }$, and V = 23.0 kV = 23.0 × 10³ V. We can use the equation:$$V = Ed$$where:* V is the potential difference between the plates in volts (V)* E is the electric field strength between the plates in volts per meter (V/m)* d is the distance between the plates in meters (m)to find the electric field strength:$$E = \frac{V}{d} = \frac{23.0 \times 10^3 \mathrm { V}}{d}$$The electric field strength is also given by:$$E = \frac{\sigma}{\epsilon_0}$$Setting these two equations equal to each other, we get:$$\frac{23.0 \times 10^3 \mathrm { V}}{d} = \frac{\pm 1.0 \times 10^{-9} \mathrm { C } / \mathrm { m } ^ { 2 }}{8.85 \times 10^{-12} \mathrm { C } ^ { 2 } / \mathrm { N } \mathrm { m } ^ { 2 }}$$Solving for d, we get:$$d = \frac{23.0 \times 10^3 \mathrm { V} \times 8.85 \times 10^{-12} \mathrm { C } ^ { 2 } / \mathrm { N } \mathrm { m } ^ { 2 }}{\pm 1.0 \times 10^{-9} \mathrm { C } / \mathrm { m } ^ { 2 }} = \pm 2.05 \times 10^{-3} \mathrm { m }$$The surface area of each plate is then:$$A = \frac{C d}{\epsilon_0} = \frac{3.0 \times 10^{-12} \mathrm { F } \times \pm 2.05 \times 10^{-3} \mathrm { m }}{8.85 \times 10^{-12} \mathrm { C } ^ { 2 } / \mathrm { N } \mathrm { m } ^ { 2 }} = \pm 69.0 \mathrm { mm } ^ { 2 }$$Therefore, the surface area of each plate is 69 mm².

A 30-PFcapacitor Consists of Two Large Closely-Spaced Parallel Plates That Have