Electrostatics and Electric Fields: Key Concepts and Applications

Electrostatics and Electric Forces

Gravitational vs. Electrical Forces on Subatomic Particles

At Earth's surface, an electron experiences a gravitational force $F_g = m_e g$. However, the electrical force between a proton and electron, given by Coulomb's law, is typically much stronger at molecular scales.

• Gravitational Force: $F_g = m_e g$

• Coulomb's Law: $F_e = k_e \frac{|q_1 q_2|}{r^2}$

• Distance for Equal Forces: To find the distance $r$ where $F_e = F_g$, solve: $r = \sqrt{\frac{k_e e^2}{m_e g}}$ For $k_e = 9.0 \times 10^9$ N·m²/C², $e = 1.60 \times 10^{-19}$ C, $m_e = 9.11 \times 10^{-31}$ kg, $g = 9.8$ m/s²: $r = \sqrt{\frac{(9.0 \times 10^9)(1.60 \times 10^{-19})^2}{(9.11 \times 10^{-31})(9.8)}} \approx 5.1$ m

• Application: On the molecular scale ($\sim 10^{-10}$ m), electrical forces vastly outweigh gravity.

Electric Fields and Forces

Force on Charges in an Electric Field

The force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$. The direction depends on the sign of the charge.

• Example: A $-1$ μC charge experiences a $10^7$ N force. For a proton ($q_p = 1.6 \times 10^{-19}$ C): $F_p = q_p \frac{F}{q} = (1.6 \times 10^{-19}) \frac{10^7}{-1.0 \times 10^{-6}} = -1.6$ pN The force on the proton is opposite in direction to that on the original charge.

Superposition Principle and Charge Arrangements

Forces Between Multiple Charges

When multiple charges are present, the net force on any charge is the vector sum of the forces from all other charges (superposition principle).

• Example: For $q_1 = 25$ μC, $q_2 = 20$ μC, and $q_3$ unknown, with forces along the $x$-axis: Use vector analysis and Coulomb's law to solve for $q_3$ and the magnitude of the force on $q_1$.

• Key Equations: $\vec{F} = k_e \frac{q_1 q_2}{r^2} \hat{r}$ Superposition: $\vec{F}_{\text{net}} = \vec{F}_{12} + \vec{F}_{13}$

Electric Field Lines and Net Charge

Interpreting Field Line Diagrams

Electric field lines indicate the direction and relative magnitude of electric fields around charges. The number of lines is proportional to the charge magnitude.

• Example: If the central charge is $-3$ μC and each outer charge is $+3$ μC, the net charge is: $+3$ μC $+3$ μC $-3$ μC $= +3$ μC

• Application: Field lines point toward negative charges and away from positive charges.

Gauss's Law and Electric Flux

Calculating Electric Flux Through a Sphere

Gauss's law relates the electric flux through a closed surface to the charge enclosed:

• Gauss's Law: $\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

• Example: For charges $+6.8$ μC and $-4.7$ μC inside a sphere: $\Phi = \frac{6.8 - 4.7}{8.85 \times 10^{-12}} = 240$ kN·m²/C

• Key Point: Electric flux depends only on the enclosed charge, not the sphere's size.

Electric Field of Spherical Charge Distributions

Field Inside and Outside a Spherical Shell

For a spherical shell with charge $Q$ and radius $R$, the electric field is:

• Inside the Shell ($r < R$): $E = 0$

• Outside the Shell ($r > R$): $E = \frac{k_e Q}{r^2}$

• Example: For a balloon of radius $70$cm and$E = 26$kN/C at the surface: At$r = 190$cm:$E = 26 \times \left(\frac{70}{190}\right)^2 = 3.5$kN/C Net charge:$Q = \frac{E R^2}{k_e} = \frac{26 \times (0.70)^2}{9.0 \times 10^9} = 1.4$ μC

Concentric Spheres: Electric Field Calculations

Field at Various Points

For a sphere of radius $10$cm with $40$ μC charge, surrounded by a shell of radius $20$ cm:

• Inside Inner Sphere ($r < 10$ cm): $E = 0$

• Between Spheres ($10$cm$< r < 20$ cm): $E = \frac{k_e Q_{\text{inner}}}{r^2}$

• Outside Both ($r > 20$ cm): $E = \frac{k_e (Q_{\text{inner}} + Q_{\text{shell}})}{r^2}$

Superposition Principle for Spherical Charge Distributions

Point Charge at Center and Shell Charge

When a point charge $q$ is at the center of a shell with charge $2q$:

• Inside Shell ($r < R$): $E = \frac{k_e q}{r^2}$

• Outside Shell ($r > R$): $E = \frac{k_e (q + 2q)}{r^2} = \frac{3 k_e q}{r^2}$

• Application: Use superposition to sum fields from both distributions.

Electric Potential and Capacitance

Energy Gained by Ions in a Potential Difference

The work done on a charge $q$ moving through a potential difference $\Delta V$ is $W = q \Delta V$.

• Example: If a particle gains $1.50$ J in $3.12$ kV, its charge is $q = \frac{W}{\Delta V} = \frac{1.50}{3120} = 4.8 \times 10^{-4}$ C

• Application: Compare to elementary charge to determine ionization state.

Capacitance of a Cell Membrane

Capacitance $C$ is defined as $C = \frac{Q}{V}$, where $Q$ is charge and $V$ is potential difference.

• Example: For $1.5 \times 10^5$ potassium ions ($1.6 \times 10^{-19}$ C each) and $V = 65$ mV: $C = \frac{1.5 \times 10^5 \times 1.6 \times 10^{-19}}{65 \times 10^{-3}} = 3.7 \times 10^{-13}$ F

• Parallel Plate Approximation: $C = \varepsilon_r \varepsilon_0 \frac{A}{d}$

Capacitors in Circuits

Equivalent Capacitance

Capacitors can be combined in series and parallel:

• Parallel: $C_{eq} = C_1 + C_2$

• Series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$

• Example: For $C_1 = 3.3$ pF, $C_2 = 2.7$ pF, $C_3 = 1.2$ pF, $C_4 = 3.3$ pF: Combine parallel and series as appropriate to find $C_{eq}$.

Current, Current Density, and Electric Field in Conductors

Safe Current in Copper Wire

Current density $J$ is $J = \frac{I}{A}$, where $I$ is current and $A$ is cross-sectional area.

• Example: For $I = 20$ A, diameter $2.1$ mm: $A = \pi (1.05 \times 10^{-3})^2 = 3.46 \times 10^{-6}$ m² $J = \frac{20}{3.46 \times 10^{-6}} = 5.78 \times 10^6$ A/m²

• Electric Field: $E = \rho J$, with $\rho$ the resistivity of copper ($1.68 \times 10^{-8}$ Ω·m)

Identifying Materials by Electrical Properties

Resistivity Calculations

Resistivity $\rho$ can be found using $\rho = \frac{RA}{L} = \frac{VA}{IL}$ for a uniform material.

• Example: For $V = 9$ V, $I = 2.6$ mA, $L = 24$ cm, $d = 2.0$ mm: $A = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}$ m² $\rho = \frac{9 \times 3.14 \times 10^{-6}}{2.6 \times 10^{-3} \times 0.24} = 0.47$ Ω·m$

• Application: This matches the resistivity of germanium.

Electric Current and Resistance in Pulsed Circuits

Taser Pulse Calculations

Current $I$ is charge per unit time, $I = \frac{Q}{\Delta t}$. Resistance $R$ is $R = \frac{V}{I}$.

• Example: For $Q = 100$ μC, $\Delta t = 0.004$ s, $V = 1200$ V: $I = \frac{100 \times 10^{-6}}{0.004} = 0.025$ A $R = \frac{1200}{0.025} = 48$ kΩ

• Application: Taser pulses are engineered to immobilize without causing permanent harm.