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: qp = +1.6\times10^{-19}\ \text{C} • Electron: qe = -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 q1, q2 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):

\mathbf F{21}=ke\,\frac{q1q2}{r^{2}}\,\hat{\mathbf r}{21} • \hat{\mathbf r}{21} is a unit vector from 1 to 2.
• Sign of q1q2 encodes attraction (negative) or repulsion (positive).

• Vector components
• Any vector \mathbf F=(Fx,Fy,Fz) has magnitude |\mathbf F| = \sqrt{Fx^{2}+Fy^{2}+Fz^{2}}

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

• Worked numerical example (from slides)
• Two equal charges q1=q2=100\,\mu\text{C}=1.0\times10^{-4}\,\text{C} separated r=0.25\,\text{m}
• Force magnitude
|\mathbf F| = ke\frac{q1q_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:

\mathbf E(\mathbf r)=k_e\frac{Q}{r^{2}}\hat{\mathbf r}

• 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 E1=(0,1.25\times10^{5})\,\text{N/C} • From –2 mC (components): \mathbf E2=(1.6\times10^{4},-1.3\times10^{4})\,\text{N/C}
• Superpose: \mathbf EP=(1.6\times10^{4},1.12\times10^{5})\,\text{N/C} • Magnitude: |\mathbf EP|=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:

\frac{mv^{2}}{r}=qE \quad\Rightarrow\quad v=\sqrt{\frac{qEr}{m}}

• 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| = ke\,|q1q2|/r^{2} • Electric field: \mathbf E = \mathbf F/q_{\text{test}} • Superposition: \mathbf E{\text{net}} = \sumi \mathbf Ei
• 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{ke |q1q_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: \PhiE = E\,4\pi r^{2}. (b) Charge magnitude via Gauss’s law: Q=\varepsilon0\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 = R0[1+\alpha (T-T0)] with \alpha=-5\times10^{-4}\,(\text{°C})^{-1}.