Comprehensive Study Notes on Electrostatics: Force Redistribution, Orbital Motion, and Charge Division Optimization

Electrostatic Force Redistribution Between Identical Spheres

• System Configuration and Initial Parameters:

  • Consider two identical small spherical conductors separated by a fixed center-to-center distance $r$.

  • Initially, both spheres carry an identical positive electric charge $Q$.

  • According to Coulomb's Law, the initial electrostatic repulsive force $F$ exerted between the two spheres in a vacuum or air is given by:     $F = \frac{K \cdot Q \cdot Q}{r^2} = \frac{K \cdot Q^2}{r^2}$     where $K = \frac{1}{4 \cdot \pi \cdot \varepsilon_0}$ is Coulomb's constant, $Q$ is the charge on each sphere, and $r$ is the separation distance between their centers.

• Charge Transfer Mechanics:

  • A charge transfer process is conducted where $50\%$ of the charge from one sphere is transferred to the other sphere.

  • Initial charge on the first sphere: $Q_1 = Q$

  • Initial charge on the second sphere: $Q_2 = Q$

  • Amount of charge transferred: $\Delta Q = 50\% \times Q = 0.5 \cdot Q = \frac{Q}{2}$

  • New charge remaining on the first sphere ($Q_1'$):     $Q_1' = Q_1 - \Delta Q = Q - \frac{Q}{2} = \frac{Q}{2}$

  • New charge accumulated on the second sphere ($Q_2'$):     $Q_2' = Q_2 + \Delta Q = Q + \frac{Q}{2} = \frac{3 \cdot Q}{2}$

• Calculation of New Electrostatic Force ($F_{\text{new}}$):

  • Substituting the modified charges $Q_1'$ and $Q_2'$ back into Coulomb's Law while keeping the distance $r$ constant:     $F_{\text{new}} = \frac{K \cdot Q_1' \cdot Q_2'}{r^2}$     $F_{\text{new}} = \frac{K \cdot \left(\frac{Q}{2}\right) \cdot \left(\frac{3 \cdot Q}{2}\right)}{r^2}$     $F_{\text{new}} = \frac{3}{4} \cdot \frac{K \cdot Q^2}{r^2}$

  • Expressing $F_{\text{new}}$ in terms of the original force $F$:     $F_{\text{new}} = \frac{3}{4} \cdot F$

• Evaluation of Options:

  • The resulting force after the $50\%$ charge transfer is $\frac{3}{4} \cdot F$

  • Corresponding option selection: Option (C) $\frac{3}{4} \cdot F$

Circular Orbital Motion of Charged Particles Under Electrostatic Attraction

• Problem Setup and Physical Model:

  • A particle of mass $m$ carrying a negative electric charge $-q_2$ revolves in a circular path of radius $r$ around a fixed positive point charge $q_1$

  • The central charge $q_1$ is stationary and serves as the center of the circular orbit.

  • The electrostatic force between the opposite charges $q_1$ and $-q_2$ is attractive and directed radially inward toward $q_1$

• Centripetal Force Equilibrium:

  • The attractive electrostatic force $F_{\text{electrostatic}}$ provides the required centripetal force $F_{\text{centripetal}}$ to sustain the circular orbit of the mass $m$ at a constant speed $v$

  • Electrostatic force equation:     $F_{\text{electrostatic}} = \frac{K \cdot q_1 \cdot q_2}{r^2}$

  • Centripetal force equation:     $F_{\text{centripetal}} = \frac{m \cdot v^2}{r}$

  • Equating centripetal force to electrostatic force:     $\frac{m \cdot v^2}{r} = \frac{K \cdot q_1 \cdot q_2}{r^2}$

• Derivation of Orbital Speed ($v$):

  • Multiplying both sides of the force equilibrium equation by $r$:     $m \cdot v^2 = \frac{K \cdot q_1 \cdot q_2}{r}$

  • Isolating $v^2$ by dividing by mass $m$:     $v^2 = \frac{K \cdot q_1 \cdot q_2}{m \cdot r}$

  • Taking the principal square root yields the orbital speed $v$:     $v = \sqrt{\frac{K \cdot q_1 \cdot q_2}{m \cdot r}}$

• Derivation of Period of Revolution ($T$):

  • The period of revolution $T$ represents the time required for the charged particle to complete one full circular path of circumference $2 \cdot \pi \cdot r$

  • Relation between time period, path circumference, and orbital speed:     $T = \frac{2 \cdot \pi \cdot r}{v}$

  • Substituting the derived expression for orbital speed $v$:     $T = \frac{2 \cdot \pi \cdot r}{\sqrt{\frac{K \cdot q_1 \cdot q_2}{m \cdot r}}}$

  • Expressing the numerator inside the radical:     $T = \sqrt{\frac{4 \cdot \pi^2 \cdot r^2}{\frac{K \cdot q_1 \cdot q_2}{m \cdot r}}}$     $T = \sqrt{\frac{4 \cdot \pi^2 \cdot m \cdot r^3}{K \cdot q_1 \cdot q_2}}$

• Final Answers for Orbital Motion:

  • Speed of revolution:     $v = \sqrt{\frac{K \cdot q_1 \cdot q_2}{m \cdot r}}$

  • Period of revolution:     $T = \sqrt{\frac{4 \cdot \pi^2 \cdot m \cdot r^3}{K \cdot q_1 \cdot q_2}}$

Maximization of Electrostatic Repulsive Force for Divided Charges

• Problem Formulation:

  • A total electric charge $Q$ is split into two constituent point charges, $q_1$ and $q_2$, such that:     $q_1 + q_2 = Q \implies q_2 = Q - q_1$

  • The two charges $q_1$ and $q_2$ are placed at a fixed distance $r$ apart.

  • The goal is to determine the optimal charge division ratio that maximizes the repulsive electrostatic force between them.

• Mathematical Expression for Electrostatic Force:

  • Let $q_1 = q$ represent the variable charge assigned to the first part.

  • The remaining charge assigned to the second part is $q_2 = Q - q$

  • According to Coulomb's Law, the electrostatic force $F$ between the charges is:     $F = \frac{K \cdot q \cdot (Q - q)}{r^2}$

  • Expanding the numerator yields:     $F = \frac{K}{r^2} \cdot (Q \cdot q - q^2)$

• Optimization via First Derivative Test:

  • To locate the value of $q$ that maximizes $F$, take the first derivative of $F$ with respect to $q$ and set it to zero ($\frac{dF}{dq} = 0$):     $\frac{dF}{dq} = \frac{d}{dq} \left[ \frac{K}{r^2} \cdot (Q \cdot q - q^2) \right] = 0$

  • Differentiating the function inside the brackets:     $\frac{d}{dq} (Q \cdot q - q^2) = Q - 2 \cdot q$

  • Equating the derivative expression to zero:     $\frac{K}{r^2} \cdot (Q - 2 \cdot q) = 0$

  • Since $\frac{K}{r^2} \neq 0$:     $Q - 2 \cdot q = 0$     $2 \cdot q = Q$     $q = \frac{Q}{2}$

• Verification of Maximum (Second Derivative Test):

  • Computing the second derivative of $F$ with respect to $q$:     $\frac{d^2F}{dq^2} = \frac{d}{dq} \left[ \frac{K}{r^2} \cdot (Q - 2 \cdot q) \right] = -\frac{2 \cdot K}{r^2}$

  • Because Coulomb's constant $K > 0$ and distance squared $r^2 > 0$, the second derivative $\frac{d^2F}{dq^2} < 0$ is strictly negative, confirming that $q = \frac{Q}{2}$ yields a absolute maximum force $F_{\text{max}}$.

• Charge Distribution and Ratio Calculations:

  • Optimum magnitudes of the divided charges:     $q_1 = q = \frac{Q}{2}$     $q_2 = Q - q = Q - \frac{Q}{2} = \frac{Q}{2}$

  • Ratio of divided charge to total charge:     $\frac{q_1}{Q} = \frac{\frac{Q}{2}}{Q} = \frac{1}{2}$

  • Ratio of total charge to divided charge:     $\frac{Q}{q} = 2$

  • Ratio between the two divided charges:     $\frac{q_1}{q_2} = \frac{\frac{Q}{2}}{\frac{Q}{2}} = 1$