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

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 \times 10^{-4} \mathrm{~m}^{2} and a plate separation d=1.00 \times 10^{-3} \mathrm{~m}. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a 3.00-\mathrm{V} 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.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) Find the capacitance.

Substitute into Equation 16.9:

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

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

Substitute into Equation 16.8:

\begin{aligned}C=\frac{Q}{\Delta V} \quad  \rightarrow \quad Q=C \Delta V &=\left(1.77 \times 10^{-12} \mathrm{~F}\right)(3.00 \mathrm{~V}) \\&=5.31 \times 10^{-12}  \mathrm{C}\end{aligned}

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

Charge density is charge divided by area:

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

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

Apply \Delta V=E d :

E=\frac{\Delta V}{d}=\frac{3.00 \mathrm{~V}}{1.00 \times 10^{-3} \mathrm{~m}}=3.00 \times 10^{3} \mathrm{~V} / \mathrm{m}

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

Related Answered Questions