A PARALLEL-PLATE CAPACITOR GOAL Calculate fundamental physical properties of a parallel-plate capacitor. PROBLEM A parallel-plate capacitor has an area A=2.00×10^−4 m² and a plate separation d=1.00×10^−3 m. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is

Question

A PARALLEL-PLATE CAPACITOR

GOAL Calculate fundamental physical properties of a parallel-plate capacitor.

PROBLEM A parallel-plate capacitor has an area [latex]A=2.00 imes 10^{-4} \mathrm{~m}^{2}[/latex] and a plate separation [latex]d=1.00 imes 10^{-3} \mathrm{~m}[/latex]. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a [latex]3.00-\mathrm{V}[/latex] battery? Calculate (c) the charge density on the positive plate, assuming the density is uniform, and (d) the magnitude of the electric field between the plates.

STRATEGY Parts (a) and (b) can be solved by substituting into the basic equations for capacitance. In part (c) use the definition of charge density, and in part (d) use the fact that the voltage difference equals the electric field times the distance.

Accepted Answer

(a) Find the capacitance.

Substitute into Equation 16.9:

[latex]\begin{aligned}&C=\epsilon_{0} \frac{A}{d}=\left(8.85 imes 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2} ight)\left(\frac{2.00 imes 10^{-4} \mathrm{~m}^{2}}{1.00 imes 10^{-3} \mathrm{~m}} ight) \&C=1.77 imes 10^{-12} \mathrm{~F}=1.77 \mathrm{pF}\end{aligned}[/latex]

(b) Find the charge on the positive plate after the capacitor is connected to a [latex]3.00-\mathrm{V}[/latex] battery.

Substitute into Equation 16.8:

[latex]\begin{aligned}C=\frac{Q}{\Delta V} \quad ightarrow \quad Q=C \Delta V &=\left(1.77 imes 10^{-12} \mathrm{~F} ight)(3.00 \mathrm{~V}) \&=5.31 imes 10^{-12} \mathrm{C}\end{aligned}[/latex]

(c) Calculate the charge density on the positive plate.

Charge density is charge divided by area:

[latex]\sigma=\frac{Q}{A}=\frac{5.31 imes 10^{-12} \mathrm{C}}{2.00 imes 10^{-4} \mathrm{~m}^{2}}=2.66 imes 10^{-8} \mathrm{C} / \mathrm{m}^{2}[/latex]

(d) Calculate the magnitude of the electric field between the plates.

Apply [latex]\Delta V=E d[/latex] :

[latex]E=\frac{\Delta V}{d}=\frac{3.00 \mathrm{~V}}{1.00 imes 10^{-3} \mathrm{~m}}=3.00 imes 10^{3} \mathrm{~V} / \mathrm{m}[/latex]

REMARKS The answer to part (d) could also have been obtained from the electric field derived for a parallel plate capacitor, Equation [latex]15.13, E=\sigma / \epsilon_{0}[/latex].

Question 16.6: A PARALLEL-PLATE CAPACITOR GOAL Calculate fundamental physic...

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