# 19.3 CAPACITORS A **capacitor** is an electronic component used to store electrical energy. A common type is the **parallel plate capacitor**, which consists of two identical, parallel conducting plates separated by a distance $d$. The insulating medium between the plates is called a **dielectric**. - **Charging:** When connected to a battery, equal and opposite charges, $+Q$ and $-Q$, accumulate on the plates. While the capacitor remains electrically neutral overall, it is said to "store" a charge $Q$. - **Charge Storage:** Due to electrostatic induction and mutual attraction, the charges remain on the inner surfaces of the plates, allowing the charge to be stored even after the battery is disconnected. ## Capacitance of a Capacitor **Definition:** Capacitance is the measure of a capacitor's ability to store charge. It is defined as the ratio of the magnitude of the charge $Q$ on one plate to the potential difference $V$ between the plates. **Formula:** The relationship between charge, voltage, and capacitance is: $$ Q = CV $$ or $$ C = \frac{Q}{V} $$ **Unit:** The SI unit of capacitance is the **farad (F)**, named after Michael Faraday. - $1 \text{ Farad} = 1 \text{ Coulomb / Volt}$ - One farad is a very large unit. Practical capacitors are typically in the range of picofarads ($1 \text{ pF} = 10^{-12} \text{ F}$) to millifarads ($1 \text{ mF} = 10^{-3} \text{ F}$). ## Capacitance of a Parallel Plate Capacitor The capacitance of a parallel plate capacitor is determined by its physical characteristics: the area of the plates, the distance between them, and the dielectric material. **Derivation:** 1. The electric field $E$ between the plates is uniform and given by $E = V/d$. 2. From Gauss's law, the electric field is also related to the surface charge density $\sigma$ by $E = \frac{\sigma}{\varepsilon_0}$, where $\sigma = Q/A$. 3. Equating the two expressions for $E$ gives: $\frac{V}{d} = \frac{Q}{A \varepsilon_0}$. 4. Rearranging for $C = Q/V$ yields the capacitance formulas. **Formulas:** | Medium | Formula | Description | | :--- | :--- | :--- | | **Vacuum/Air** | $C_{vac} = \frac{A \varepsilon_0}{d}$ | $A$ = area of each plate, $d$ = separation, $\varepsilon_0$ = permittivity of free space ($8.85 \times 10^{-12} \text{ F/m}$). | | **Dielectric** | $C_{med} = \frac{A \varepsilon_0 \varepsilon_r}{d}$ | $\varepsilon_r$ is the relative permittivity (or dielectric constant) of the insulating material. | ## Relative Permittivity ($\varepsilon_r$) **Definition:** The ratio of the capacitance of a capacitor with a dielectric medium to its capacitance in a vacuum. It is a dimensionless factor that indicates how much a dielectric material can increase capacitance. $$ \varepsilon_r = \frac{C_{med}}{C_{vac}} $$ **Effect:** Inserting a dielectric between the plates enhances the capacitor's ability to store charge, thus increasing its capacitance. This occurs due to **Electric Polarization**. ## Worked Examples **Q1:** What is the capacitance of a parallel plate capacitor with metal plates, each of area $1.00 \text{ m}^2$, separated by 1.00 mm? **A:** Using the formula $C = A\varepsilon_0/d$: $$ C = \frac{(1.00 \text{ m}^2)(8.85 \times 10^{-12} \text{ F/m})}{(1.00 \times 10^{-3} \text{ m})} $$ $$ C = 8.85 \times 10^{-9} \text{ F} = 8.85 \text{ nF} $$ **Q2:** What charge is stored in the capacitor above if a voltage of $3.00 \times 10^3 \text{ V}$ is applied to it? **A:** Using the formula $Q = CV$: $$ Q = (8.85 \times 10^{-9} \text{ F})(3.00 \times 10^3 \text{ V}) = 26.55 \times 10^{-6} \text{ C} = 26.6 \mu\text{C} $$ **Q3:** What is the capacitance of a large Van de Graaff generator's terminal, given that it stores $8.00 \text{ mC}$ of charge at a voltage of $12.0 \text{ mV}$? **A:** Using the formula $C = Q/V$: $$ C = \frac{8.00 \times 10^{-3} \text{ C}}{12.0 \times 10^{-3} \text{ V}} = 0.667 \text{ F} $$ - Capacitors are components designed to store electrical energy. They consist of two conductors separated by an insulator (dielectric). - Capacitance ($C$) is the ratio of stored charge ($Q$) to the potential difference ($V$) and is measured in Farads (F). - The capacitance of a parallel plate capacitor is directly proportional to the plate area ($A$) and inversely proportional to the separation distance ($d$). - Inserting a dielectric material with relative permittivity $\varepsilon_r$ between the plates increases the capacitance. | Formula | Description | | :--- | :--- | | $C = \frac{Q}{V}$ | General definition of capacitance. | | $C_{vac} = \frac{A \varepsilon_0}{d}$ | Capacitance of a parallel plate capacitor in a vacuum. | | $C_{med} = \frac{A \varepsilon_0 \varepsilon_r}{d}$ | Capacitance of a parallel plate capacitor with a dielectric. | | $\varepsilon_r = \frac{C_{med}}{C_{vac}}$ | Definition of relative permittivity. | Capacitors are fundamental components in countless electronic circuits, used for filtering, timing, and energy storage. For further study, see and .