Physics 2

Electrostatic Forces and Vector Coulomb's Law

  • The electrostatic force between two stationary point charges is governed by Coulomb's Law, expressed in vector form to account for both magnitude and directional orientation.

  • Coulomb Constant (kek_e):

    • Value: ke=9×109 N×m2/C2k_e = 9 \times 10^9\text{ N}\times\text{m}^2/\text{C}^2

  • Fundamental Quantities:

    • qi,qjq_i, q_j: Magnitudes of point charges measured in Coulombs (C\text{C}).

    • ri,rjr_i, r_j: Position vectors of charge ii and charge jj in space.

    • dijd_{ij}: Displacement vector from position ii to position jj.

    • ∥dij∥\|d_{ij}\|: Distance (scalar magnitude) separating charge ii and charge jj

    • d^ij\hat{d}_{ij}: Unit vector pointing in the direction from charge ii to charge jj

  • Nature of Electrostatic Forces:

    • Like charges (charges of the same sign) produce a positive product qiqj>0q_i q_j > 0, resulting in a repulsive force.

    • Opposite charges (charges of different signs) produce a negative product qiqj<0q_i q_j < 0, resulting in an attractive force.

Vector Mechanics and Geometric Framework

  • To formulate the electrostatic force FijF_{ij} exerted by source charge ii on target charge jj, a strict vector orientation framework must be established.

  • Vector Addition Relationship:

    • Given position vectors rir_i and rjr_j, the displacement vector dijd_{ij} going from charge ii to charge jj satisfies the vector addition triangle:   ri+dij=rjr_i + d_{ij} = r_j

    • Re-arranging gives the displacement vector:   dij=rj−rid_{ij} = r_j - r_i

    • The subtraction index order reverses the direction notation: To determine displacement from position ii to position jj, subtract the initial position vector rir_i from the target position vector rjr_j

  • Magnitude Calculation:

    • For two-dimensional position coordinates ri=(xi,yi)r_i = (x_i, y_i) and rj=(xj,yj)r_j = (x_j, y_j), the displacement vector components are:   dij=(xj−xi,yj−yi)d_{ij} = (x_j - x_i, y_j - y_i)

    • The magnitude (distance) is:   ∥dij∥=(xj−xi)2+(yj−yi)2\|d_{ij}\| = \sqrt{(x_j - x_i)^2 + (y_j - y_i)^2}

    • Note that magnitude is symmetric: ∥dij∥=∥dji∥\|d_{ij}\| = \|d_{ji}\|

  • Unit Direction Vector (d^ij\hat{d}_{ij}):

    • Any vector is defined as its magnitude multiplied by its unit direction vector:   dij=∥dij∥d^ijd_{ij} = \|d_{ij}\| \hat{d}_{ij}

    • Isolating the unit direction vector yields:   d^ij=dij∥dij∥=rj−ri∥rj−ri∥\hat{d}_{ij} = \frac{d_{ij}}{\|d_{ij}\|} = \frac{r_j - r_i}{\|r_j - r_i\|}

  • General Vector Coulomb's Law Formula:

    • The force FijF_{ij} exerted by charge ii on charge jj is given by:   Fij=keqiqj∥dij∥2d^ijF_{ij} = k_e \frac{q_i q_j}{\|d_{ij}\|^2} \hat{d}_{ij}

    • Substituting the expression for the unit direction vector gives:   Fij=keqiqj∥dij∥3dijF_{ij} = k_e \frac{q_i q_j}{\|d_{ij}\|^3} d_{ij}

Comprehensive Example 1: Force on an Attractive Opposite Charge

  • System Configuration:

    • Charge 1: q1=−3 Cq_1 = -3\text{ C} located at position vector r1=(−4 m,3 m)r_1 = (-4\text{ m}, 3\text{ m})

    • Charge 2: q2=2 Cq_2 = 2\text{ C} located at position vector r2=(2 m,−2 m)r_2 = (2\text{ m}, -2\text{ m})

    • Target Problem: Calculate the net electrostatic force F1F_1 (which is the force F21F_{21} exerted by charge 2 on charge 1).

  • Step-by-Step Calculation:

    • Step 1: Compute Charge Product     q1q2=(−3 C)×(2 C)=−6 C2q_1 q_2 = (-3\text{ C}) \times (2\text{ C}) = -6\text{ C}^2

    • Step 2: Calculate Displacement Vector (d21d_{21})

    • Since the target charge is q1q_1 and the source charge is q2q_2, displacement vector d21d_{21} points from position 2 to position 1:     d21=r1−r2d_{21} = r_1 - r_2     d21=(−4,3)−(2,−2)=(−4−2,3−(−2))=(−6,5)d_{21} = (-4, 3) - (2, -2) = (-4 - 2, 3 - (-2)) = (-6, 5)

    • Step 3: Calculate Displacement Magnitude (∥d21∥\|d_{21}\|)     ∥d21∥=(−6)2+52=36+25=61 m\|d_{21}\| = \sqrt{(-6)^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}\text{ m}

    • Step 4: Formulate Direction Vector (d^21\hat{d}_{21})     d^21=d21∥d21∥=(−6,5)61\hat{d}_{21} = \frac{d_{21}}{\|d_{21}\|} = \frac{(-6, 5)}{\sqrt{61}}

    • Step 5: Calculate Electrostatic Force (F21F_{21})     F21=keq1q2∥d21∥2d^21F_{21} = k_e \frac{q_1 q_2}{\|d_{21}\|^2} \hat{d}_{21}     F21=(9×109)−6(61)2(−6,5)61F_{21} = (9 \times 10^9) \frac{-6}{(\sqrt{61})^2} \frac{(-6, 5)}{\sqrt{61}}     F21=(9×109)−661(−6,5)61F_{21} = (9 \times 10^9) \frac{-6}{61} \frac{(-6, 5)}{\sqrt{61}}     F21=−54×1096161(−6,5)F_{21} = \frac{-54 \times 10^9}{61 \sqrt{61}} (-6, 5)

    • Step 6: Component Numerical Evaluation

    • Evaluated scalar denominator: 6161≈61×7.81025=476.4361 \sqrt{61} \approx 61 \times 7.81025 = 476.43

    • xx-component:       F21,x=−54×(−6)476.43×109=324476.43×109≈0.68×109 NF_{21,x} = \frac{-54 \times (-6)}{476.43} \times 10^9 = \frac{324}{476.43} \times 10^9 \approx 0.68 \times 10^9\text{ N}

    • yy-component:       F21,y=−54×5476.43×109=−270476.43×109≈−0.57×109 NF_{21,y} = \frac{-54 \times 5}{476.43} \times 10^9 = \frac{-270}{476.43} \times 10^9 \approx -0.57 \times 10^9\text{ N}

    • Vector Force Result:       F1=(0.68×109 N,−0.57×109 N)F_1 = (0.68 \times 10^9\text{ N}, -0.57 \times 10^9\text{ N})

  • Directional Analysis & Geometric Verification:

    • The calculated force on charge 1 has a positive xx-component (+0.68×109 N+0.68 \times 10^9\text{ N}) and a negative yy-component (−0.57×109 N-0.57 \times 10^9\text{ N}).

    • This vector points down and to the right, directly toward charge 2.

    • Because q1q_1 (−3 C-3\text{ C}) and q2q_2 (+2 C+2\text{ C}) are opposite in sign, the force is attractive, pulling charge 1 directly toward charge 2, confirming physical accuracy.

Comprehensive Example 2: Force on a Repulsive Like Charge

  • System Configuration:

    • Charge 1: q1=1 Cq_1 = 1\text{ C} located at position vector r1=(4 m,3 m)r_1 = (4\text{ m}, 3\text{ m})

    • Charge 2: q2=2 Cq_2 = 2\text{ C} located at position vector r2=(4 m,−3 m)r_2 = (4\text{ m}, -3\text{ m})

    • Target Problem: Calculate the net electrostatic force F1F_1 (force F21F_{21} exerted by charge 2 on charge 1).

  • Step-by-Step Calculation:

    • Step 1: Compute Charge Product     q1q2=(1 C)×(2 C)=2 C2q_1 q_2 = (1\text{ C}) \times (2\text{ C}) = 2\text{ C}^2

    • Step 2: Calculate Displacement Vector (d21d_{21})     d21=r1−r2=(4,3)−(4,−3)=(4−4,3−(−3))=(0,6)d_{21} = r_1 - r_2 = (4, 3) - (4, -3) = (4 - 4, 3 - (-3)) = (0, 6)

    • Step 3: Calculate Displacement Magnitude (∥d21∥\|d_{21}\|)     ∥d21∥=02+62=36=6 m\|d_{21}\| = \sqrt{0^2 + 6^2} = \sqrt{36} = 6\text{ m}

    • Step 4: Formulate Unit Direction Vector (d^21\hat{d}_{21})     d^21=(0,6)6=(0,1)=y^\hat{d}_{21} = \frac{(0, 6)}{6} = (0, 1) = \hat{y}

    • Step 5: Compute Force Vector (F21F_{21})     F21=keq1q2∥d21∥2d^21F_{21} = k_e \frac{q_1 q_2}{\|d_{21}\|^2} \hat{d}_{21}     F21=(9×109)(1)(2)62(0,1)F_{21} = (9 \times 10^9) \frac{(1)(2)}{6^2} (0, 1)     F21=(9×109)236(0,1)F_{21} = (9 \times 10^9) \frac{2}{36} (0, 1)     F21=18×10936(0,1)=0.5×109(0,1) NF_{21} = \frac{18 \times 10^9}{36} (0, 1) = 0.5 \times 10^9 (0, 1)\text{ N}     F21=(0,0.5×109 N)=0.5×109y^ NF_{21} = (0, 0.5 \times 10^9\text{ N}) = 0.5 \times 10^9 \hat{y}\text{ N}

  • Directional Analysis & Geometric Verification:

    • Both charges are positive (+1 C+1\text{ C} and +2 C+2\text{ C}), indicating a repulsive force.

    • The calculated force vector acts entirely in the positive yy-direction (y^\hat{y}), pushing charge 1 straight upward away from charge 2.

    • Correspondingly, the equal and opposite force on charge 2 (F12F_{12}) pushes charge 2 straight downward in the negative yy-direction (−y^-\hat{y}).

The Principle of Superposition for Multiple Point Charges

  • Definition:

    • The Principle of Superposition states that when multiple point charges act on a target charge, the total electrostatic force exerted on the target charge is equal to the vector sum of individual forces exerted independently by each charge.

  • Mathematical Formulation:

    • For a system of nn point charges, the net electrostatic force FjF_j acting on charge jj is given by:   Fj=∑i≠jFij=∑i≠jkeqiqj∥dij∥2d^ijF_j = \sum_{i \neq j} F_{ij} = \sum_{i \neq j} k_e \frac{q_i q_j}{\|d_{ij}\|^2} \hat{d}_{ij}

  • Three-Charge Configuration:

    • For charges q1,q2,q3q_1, q_2, q_3, the total force acting on charge q3q_3 is:   F3=F13+F23F_3 = F_{13} + F_{23}

    • F13F_{13}: Force exerted by charge 1 on charge 3.

    • F23F_{23}: Force exerted by charge 2 on charge 3.

  • Four-Charge Configuration:

    • For charges q1,q2,q3,q4q_1, q_2, q_3, q_4, the total force acting on target charge q4q_4 is:   F4=F14+F24+F34F_4 = F_{14} + F_{24} + F_{34}

    • Expanded vector equation:   F4=keq1q4∥d14∥2d^14+keq2q4∥d24∥2d^24+keq3q4∥d34∥2d^34F_4 = k_e \frac{q_1 q_4}{\|d_{14}\|^2} \hat{d}_{14} + k_e \frac{q_2 q_4}{\|d_{24}\|^2} \hat{d}_{24} + k_e \frac{q_3 q_4}{\|d_{34}\|^2} \hat{d}_{34}

    • Associated displacement vectors:

    • d14=r4−r1d_{14} = r_4 - r_1

    • d24=r4−r2d_{24} = r_4 - r_2

    • d34=r4−r3d_{34} = r_4 - r_3

    • Associated unit direction vectors:

    • d^14=d14∥d14∥\hat{d}_{14} = \frac{d_{14}}{\|d_{14}\|}

    • d^24=d24∥d24∥\hat{d}_{24} = \frac{d_{24}}{\|d_{24}\|}

    • d^34=d34∥d34∥\hat{d}_{34} = \frac{d_{34}}{\|d_{34}\|}

Administrative Policies and Upcoming Concepts

  • Attendance Record Policy:

    • Attendance tracking is mandatory.

    • Absences are tracked and penalties will be assessed against student performance metrics for unexcused absences.

    • Systematic recording prevents student administrative purging, as re-enrolling purged students causes administrative complexity.

  • Upcoming Physics Concepts:

    • The next core physical concept to be developed following two-body forces and superposition is the Electric Field.