Coulomb's Law and Electrostatic Principles Study Notes

According to Coulomb's Law, the electrostatic force ( $F$ ) between two point charges separated by a distance ( $r$ ) is defined by two primary proportionalities: - (i) Directly proportional to the product of their magnitudes: The force increases as the magnitude of the charges increases. Mathematically, $F \propto q_1 q_2$ . - (ii) Inversely proportional to the square of the distance between them: The force decreases rapidly as the distance between the point charges increases. Mathematically, $F \propto \frac{1}{r^2}$ .

Consider two point charges, $q_1$ and $q_2$ , separated by a distance $r$ , with $F$ representing the electrostatic force between them.
By combining the proportionalities: - $F \propto \frac{q_1 q_2}{r^2}$ - $F = k \frac{q_1 q_2}{r^2}$
The value of $k$ (the electrostatic constant) depends on the system of units used: - In S.I. Units: $k = \frac{1}{4 \pi \epsilon_0}$ - In C.G.S. Units: $k = 1$

The direction and nature of the force (attraction or repulsion) depend on the sign of the product of the charges ( $q_1 q_2$ ): - Like Charges (Repulsive Force): If $q_1 q_2 > 0$ , the force is repulsive. This occurs when both charges are positive ( $+$ and $+$ ) or both are negative ( $-$ and $-$ ). - Unlike Charges (Attractive Force): If $q_1 q_2 < 0$ , the force is attractive. This occurs when one charge is positive and the other is negative ( $+$ and $-$ ).

$\epsilon_0$ (Permittivity of Free Space): - Numerical value: $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 \, N^{-1} \, m^{-2}$ - Dimensional formula: $[M^{-1} L^{-3} T^4 A^2]$
$k$ (Electrostatic Constant): - Numerical value (in vacuum/air): $k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, N \, m^2 \, C^{-2}$ - Dimensional formula: $[M L^3 T^{-4} A^{-2}]$

Scenario: A force $F$ exists between two charges $2a$ and $4a$ at distance $r$ . Calculate the new force $F'$ if the distance is halved ( $r/2$ ).
Initial Force: $F = k \frac{(2a)(4a)}{r^2}$
Final Force ( $F'$ ): - $F' = k \frac{(2a)(4a)}{(r/2)^2}$ - $F' = k \frac{8a^2}{r^2/4}$ - $F' = 4 \times \left( k \frac{8a^2}{r^2} \right)$ - $F' = 4F$

Scenario: Identical bodies A, B, and C are used. Initially, Body A has a charge of $12 \, C$ .
Process: - Body A touches identical neutral body C: The total charge ( $12 \, C + 0 \, C$ ) is shared equally. A and C now individually possess $6 \, C$ each. - Calculation of force between A and B after touching: If A has $6 \, C$ and B has a charge of $-1 \, C$ (as per calculation context) at a distance of $r = 1 \, m$ : - $F = k \frac{|q_A q_B|}{r^2}$ - $F = (9 \times 10^9) \frac{(6 \times 1)}{1^2}$ - $F = 54 \times 10^9 \, N$

Problem: Find the minimum possible electrostatic force between two charges $q_1$ and $q_2$ at a distance $r$ .
Principle: The force is directly proportional to the product of charges ( $F \propto q_1 q_2$ ). To minimize the force, the charges must be at their minimum possible magnitude.
Minimum Charge: The smallest independent charge is the elementary charge ( $e$ ).
Result: - $q_1 = q_2 = e$ - $F_{min} = k \frac{e^2}{r^2} \, N$

Scenario: Two charges $e$ and $4e$ are separated by a distance $a$ . Find the position $x$ of a third charge $q$ such that it is in equilibrium (Net force = 0).
Logic: The third charge must be placed between the two charges.
Calculation: - Let $x$ be the distance from charge $e$ . Then the distance from $4e$ is $(a - x)$ . - $k \frac{q e}{x^2} = k \frac{q (4e)}{(a-x)^2}$ - $\frac{1}{x^2} = \frac{4}{(a-x)^2}$ - Taking square root: $\frac{1}{x} = \frac{2}{a-x}$ - $a - x = 2x$ - $3x = a$ - $x = \frac{a}{3}$
Note: The position from the $4e$ charge would be $a - x = a - a/3 = 2a/3$ .

Scenario: Two charges $4e$ and $-e$ are separated by distance $a$ . Find the position of the third charge for equilibrium.
Logic: For opposite charges, the null point (net force = 0) lies outside the charges, on the side of the charge with the smaller magnitude.
Calculation: - Let the third charge be at distance $x$ from the $-e$ charge (and thus $a+x$ from the $4e$ charge). - $k \frac{4e}{(a+x)^2} = k \frac{e}{x^2}$ - $\frac{4}{(a+x)^2} = \frac{1}{x^2}$ - Taking square root: $\frac{2}{a+x} = \frac{1}{x}$ - $2x = a + x$ - $x = a$
Conclusion: The null point is located at a distance $a$ from the $-e$ charge, outside the segment joining the two charges.