Lectures I & II Notes – Electric Charge, Force & Field

Electric Charge

• Intrinsic property of matter carried by sub-atomic particles
• Exists in two kinds: positive (+) and negative (–)
• Fundamental carriers in normal matter
• Proton: $q<em>p = +1.6\times10^{-19}\ \text{C}$ • Electron: $q</em>e = -1.6\times10^{-19}\ \text{C}$

• Unit of charge is the coulomb (C)
• Historical note: named after Charles‐Augustin de Coulomb (1756–1806)

• Sub-structure insight
• Protons and neutrons are composed of quarks
• Up-type quark: $+\tfrac23e$
• Down-type quark: $-\tfrac13e$
• Electron currently considered fundamental (no smaller constituents known)

• Conservation of charge
• Total charge in an isolated system remains constant; it can be redistributed but not created or destroyed.

• Like charges repel, unlike charges attract
• The interaction is described quantitatively by Coulomb’s force law (next section).

• Everyday demonstration — balloon & hair
• Rubbing a balloon on hair transfers electrons from hair to balloon (frictional charging).
• Results
• Hair: positive, like strands repel → hair stands up
• Balloon: negative, is attracted to both hair and a neutral wall.
• Near the wall, balloon repels wall electrons ⇒ wall surface becomes momentarily positive ⇒ attraction (induced charge separation).

Coulomb’s Law – Electrostatic Force

• Magnitude between two point charges $q<em>1, q</em>2$ separated by distance $r$:

|\mathbf F| = ke \frac{|q1 q2|}{r^{2}}, \qquad ke = 9\times10^{9}\ \text{N·m}^2\text{/C}^2

• Vector form (force on 2 due to 1):
$<br /> \mathbf F<em>{21}=k</em>e\,\frac{q<em>1q</em>2}{r^{2}}\,\hat{\mathbf r}<em>{21}$ • $\hat{\mathbf r}</em>{21}$ is a unit vector from 1 to 2.
• Sign of $q<em>1q</em>2$ encodes attraction (negative) or repulsion (positive).

• Vector components
• Any vector $\mathbf F=(F<em>x,F</em>y,F<em>z)$ has magnitude $|\mathbf F| = \sqrt{F</em>x^{2}+F<em>y^{2}+F</em>z^{2}}<br />$

• Superposition principle
• Net force on a charge equals the vector sum of forces exerted by all other charges:
$\mathbf F<em>{\text{net}} = \sum</em>{i} \mathbf F_i$

• Worked numerical example (from slides)
• Two equal charges $q<em>1=q</em>2=100\,\mu\text{C}=1.0\times10^{-4}\,\text{C}$ separated $r=0.25\,\text{m}$
• Force magnitude
$|\mathbf F| = k<em>e\frac{q</em>1q_2}{r^{2}} = 9\times10^{9}\,\frac{(10^{-4})^{2}}{0.25^{2}} = 1.44\times10^{3}\,\text{N}$
• Equivalent weight: same as mass $m=\tfrac{|\mathbf F|}{g}\approx\tfrac{1440}{9.8}\approx147\,\text{kg‐force}$ (illustrates strength of electrostatic forces).

• Three-body example (electron between two protons)
• Protons separated $3.6\,\text{nm}$; electron sits $1.2\,\text{nm}$ from one proton.
• Using superposition one obtains net force $|\mathbf F| \approx 0.12\,\text{nN}$ toward the nearest proton.

Electric Field $\mathbf E$

• Defined at any point in space as force per unit positive test charge:
$\mathbf E = \frac{\mathbf F}{q_{\text{test}}}$ with units $\text{N/C}$ or equivalently $\text{V/m}$.

• For a single point charge $Q$:
$<br /> \mathbf E(\mathbf r)=k_e\frac{Q}{r^{2}}\hat{\mathbf r}<br />$
• Direction is radially outward if Q>0, inward if Q<0.

• Field-line representation
• Lines start on positive charges and terminate on negative charges.
• Density of lines ∝ magnitude of field.
• Tangent to a line gives the direction of $\mathbf E$.
• A positive test charge follows the direction of arrows.

• Field between two charges
• Unlike pair: lines go from + to –.
• Like pair: lines repel, creating a mid-plane of zero field.

• Field between parallel plates (large, uniformly charged conductors)
• Lines are parallel & equally spaced ⇒ uniform field
• Magnitude: $|\mathbf E| = \sigma/\varepsilon_0$ (derivation in capacitor unit).

• Example (two point charges and point P)
• Charges: $+5.0\,\text{mC}$ at origin, $-2.0\,\text{mC}$ at $x=0.74\,\text{m}$.
• Point $P$ at $(0,0.60)\,\text{m}$.
• Field contributions
• From +5 mC (pure $\hat y$): $\mathbf E<em>1=(0,1.25\times10^{5})\,\text{N/C}$ • From –2 mC (components): $\mathbf E</em>2=(1.6\times10^{4},-1.3\times10^{4})\,\text{N/C}$
• Superpose: $\mathbf E<em>P=(1.6\times10^{4},1.12\times10^{5})\,\text{N/C}$ • Magnitude: $|\mathbf E</em>P|=1.13\times10^{5}\,\text{N/C}$.
• If $q_{\text{test}}=+1.5\,\text{mC}$ placed at P, force
$\mathbf F=q\mathbf E = 1.5\times10^{-3}\,\text{C}\,\times1.13\times10^{5}\,\text{N/C}=0.51\,\text{N}.$

Electric Dipole

• Dipole: two equal and opposite charges separated by distance $\ell$
• Dipole moment vector: $\mathbf p = q\,\ell\,\hat{\mathbf d}$ (from – to +).
• In a uniform $\mathbf E$ field: net force zero but torque
$\tau = pE\sin\theta = |\mathbf p \times \mathbf E|$.

• Water molecule (H$_2$O) is a natural permanent dipole ⇒ explains many solvent properties.

Electrostatic Analyzer (Velocity Selector)

• Charged particle of mass $m$ and charge $q$ enters region with uniform $\mathbf E$; force $\mathbf F=q\mathbf E$ ⇒ acceleration $a=qE/m$.

• For curved‐plate analyzer (radius $r$) selecting speed $v$:
$<br /> \frac{mv^{2}}{r}=qE \quad\Rightarrow\quad v=\sqrt{\frac{qEr}{m}}<br />$
• Device permits only particles with chosen velocity to follow circular path and exit through slit.

Materials: Conductors, Insulators, Semiconductors

• Conductors (metals)
• Outer electrons delocalized (conduction electrons) ⇒ free to move.
• Good electrical & thermal conductors (electron sea also transports heat).

• Insulators (non-metals: glass, rubber, plastics)
• Electrons tightly bound to atoms ⇒ negligible current.
• Useful for preventing unwanted charge flow.

• Semiconductors (Si, Ge)
• Conductivity between that of metals and insulators; can be engineered via doping.

Recap of Key Equations

• Coulomb force (magnitude): $|\mathbf F| = k<em>e\,|q</em>1q<em>2|/r^{2}$ • Electric field: $\mathbf E = \mathbf F/q_{\text{test}}$ • Superposition: $\mathbf E</em>{\text{net}} = \sum<em>i \mathbf E</em>i$
• Dipole torque: $\boldsymbol\tau = \mathbf p \times \mathbf E$
• Force in uniform field: $\mathbf F = q\mathbf E$
• Acceleration: $a = qE/m$

Practice/Assessment Questions (from slides)

1. Two $0.5\,\text{kg}$ spheres carry $+60\,\mu\text{C}$ and $-60\,\mu\text{C}$. If $|\mathbf F|=1400\,\text{N}$, find separation $r$:
   $r = \sqrt{\dfrac{k<em>e |q</em>1q_2|}{|\mathbf F|}}.$

2. A point charge produces $E=1.25\times10^{6}\,\text{N/C}$ at $r=0.150\,\text{m}$.
   (a) Electric flux through Gaussian sphere: $\Phi<em>E = E\,4\pi r^{2}$. (b) Charge magnitude via Gauss’s law: $Q=\varepsilon</em>0\Phi_E$.

3. Parallel-plate air capacitor: plates area $A=6.80\,\text{cm}^2$ store $Q=240\,\text{pC}$ at $\Delta V = 42.0\,\text{V}$.
   • Capacitance $C=Q/\Delta V$.
   • For parallel plates: $C=\varepsilon_0 A/d$ ⇒ solve for separation $d$.
   • If $d$ doubles, $C$ halves; to store same $Q$ require $\Delta V$ doubled.

4. Temperature-dependent resistance of carbon:
   $R = R<em>0[1+\alpha (T-T</em>0)]$ with $\alpha=-5\times10^{-4}\,(\text{°C})^{-1}$.