Honors Physics Unit 18 Review: Static Electricity

Charge comes in two types, positive and negative. Like charges repel and opposite charges attract. The net charge of an ordinary object is a whole-number multiple of the elementary charge $e = 1.602 \times 10^{-19}$ C. Protons carry $+e$, electrons carry $-e$, and neutrons carry no charge.

Objects get charged by gaining or losing electrons. Protons stay locked in nuclei, so a positive object has lost electrons. It has not gained protons. Conservation of charge says the total charge of an isolated system never changes. Charge only moves around.

You can charge an object three ways. Friction rubs electrons from one material onto another, like a balloon on hair. Conduction shares charge through contact. Induction uses a nearby charged object to shove electrons to one side of a conductor, and then a ground connection lets electrons flow in or out, leaving a charge opposite to the nearby object.

Charged objects can even attract neutral ones through polarization. A charged balloon pulls the opposite charges in a wall slightly closer, and the closer charges feel a stronger pull.

Worked example. A metal sphere with $+6\,\mu\text{C}$ touches an identical sphere with $-2\,\mu\text{C}$. The total is $+4\,\mu\text{C}$, and identical conductors split it evenly, so each ends with $+2\,\mu\text{C}$. The first sphere dropped by $4\,\mu\text{C}$, so it gained $\frac{4 \times 10^{-6}}{1.602 \times 10^{-19}} \approx 2.5 \times 10^{13}$ electrons.

Coulomb's law gives the size of the electric force between two point charges:

$F = k\frac{|q_1||q_2|}{r^2}$

For charges in a vacuum, $k = 8.99 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$. The charges are in coulombs, and $r$ is their separation in meters. Use absolute values to get the magnitude, then decide direction from the signs. Like charges push apart and opposite charges pull together, always along the line connecting them.

Many test questions are about proportions instead of plugging in numbers. Doubling one charge doubles the force. Doubling both charges makes it 4 times bigger. Doubling the distance makes the force $\frac{1}{4}$ as big, and tripling it makes the force $\frac{1}{9}$ as big.

When three or more charges are present, find each pair's force separately and add them as vectors. This is the superposition principle.

Worked example. A $+3.0\,\mu\text{C}$ charge sits $0.20$ m from a $-4.0\,\mu\text{C}$ charge. Then $F = \frac{(8.99 \times 10^9)(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.20)^2} = \frac{0.108}{0.040} \approx 2.7\ \text{N}$. The signs are opposite, so the force is attractive. Each charge feels 2.7 N toward the other, matching Newton's third law.

The electric field at a point is the force per unit charge that a small positive test charge would feel there:

$E = \frac{F}{q}$

Its units are N/C. For a point charge in a vacuum, $E = \frac{k|q|}{r^2}$, so the field follows the same inverse-square pattern as Coulomb's law. The field points away from positive charges and toward negative charges.

Once you know the field, the force on any charge placed there is $F = qE$. A positive charge feels a force in the same direction as the field. A negative charge, like an electron, feels a force in the opposite direction.

Field line diagrams show the pattern. Lines start on positive charges or at infinity and end on negative charges or at infinity. They never cross, and tightly packed lines mean a stronger field. Between two equal positive charges the lines bend away from each other, and at the exact midpoint the two fields cancel so $E = 0$. Between oppositely charged parallel plates, the field is approximately uniform away from the edges.

Worked example. Find the field $0.30$ m from a $+2.0$ nC charge. $E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-9})}{(0.30)^2} = \frac{17.98}{0.090} \approx 200\ \text{N/C}$, pointing away from the charge. An electron placed there feels $F = (1.6 \times 10^{-19})(200) = 3.2 \times 10^{-17}$ N, pointing toward the charge.

Electric potential energy works like gravitational potential energy. A charge has stored energy because of its position in a field. For two point charges in a vacuum, choose $U = 0$ at infinite separation. Then $U = \frac{kq_1q_2}{r}$, with the signs included. Like charges give positive $U$, and opposite charges give negative $U$.

Electric potential is that energy divided by the charge, $V = \frac{U}{q}$. Its unit is the volt, and 1 V = 1 J/C. For a point charge in a vacuum, choosing $V = 0$ at infinity gives $V = \frac{kq}{r}$. Potential is a scalar, so to find the total potential from several charges you just add the numbers with their signs and skip the vector work.

What matters physically is potential difference, often called voltage. When a charge moves through a potential difference, the energy change is $\Delta U = q\Delta V$. The electric force pushes positive charges toward lower potential and negative charges toward higher potential. A charge can move in a different direction if it already has velocity or other forces act on it. Field lines point toward lower potential. In a uniform field, the magnitude of the potential difference across a distance $d$ along the field is $|\Delta V| = Ed$, so field strength can also be written in V/m.

Worked example. A proton starts at rest and accelerates through a potential drop of 100 V. With $\Delta V = -100$ V, the field does work $W = -q\Delta V = -(1.6 \times 10^{-19})(-100) = 1.6 \times 10^{-17}$ J. If no other forces do work, that becomes kinetic energy, so the proton ends with $1.6 \times 10^{-17}$ J, or 100 electron-volts.

A capacitor is two conductors separated by an insulator. Connect it to a battery and one plate gets $+Q$ while the other gets $-Q$. Capacitance is the charge stored per volt:

$C = \frac{Q}{V}$

The unit is the farad (F), which is huge, so real capacitors are usually in microfarads or picofarads. For parallel plates, $C = \frac{\kappa \varepsilon_0 A}{d}$, where $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m. Bigger plates mean more capacitance, and a smaller gap also means more capacitance.

The stored energy is $U = \frac{1}{2}CV^2$, which you can also write as $U = \frac{1}{2}QV$ or $U = \frac{Q^2}{2C}$.

A dielectric is an insulating material placed between the plates. It polarizes in the electric field. With plate charge fixed, this polarization reduces the field and voltage. With voltage fixed, more charge flows onto the plates instead. When the dielectric fills the gap, the dielectric constant $\kappa$ gives the factor $C = \kappa C_0$, where $C_0$ is the capacitance with a vacuum in the gap. Air has $\kappa \approx 1$.

Worked example. Air-filled plates have area $0.020\ \text{m}^2$ and are $1.0$ mm apart. $C = \frac{(8.85 \times 10^{-12})(0.020)}{0.0010} = 1.77 \times 10^{-10}$ F, or 177 pF. At 12 V, $Q = CV = (1.77 \times 10^{-10})(12) \approx 2.1 \times 10^{-9}$ C and $U = \frac{1}{2}(1.77 \times 10^{-10})(12)^2 \approx 1.3 \times 10^{-8}$ J. Filling the gap with a $\kappa = 3$ plastic raises $C$ to 531 pF.