Hello, aspiring engineers! Welcome to this deep dive into one of the foundational concepts of electrostatics:
Coulomb's Law and the
Superposition Principle. These principles are the bedrock upon which our understanding of electric fields and potentials is built, and mastering them is absolutely critical for both CBSE exams and, more importantly, for cracking the IIT JEE. So, let's roll up our sleeves and explore!
1. Unveiling Coulomb's Law: The Force Between Charges
We've already established that like charges repel and unlike charges attract. But how *much* force do they exert on each other? This question was elegantly answered by the French physicist Charles-Augustin de Coulomb in 1785, leading to what we now know as
Coulomb's Law.
Coulomb's Law quantifies the electrostatic force between two stationary point charges. A "point charge" is an idealized charge concentrated at a single point in space, much like a "point mass" in mechanics.
1.1. Statement of Coulomb's Law
The law states:
- The electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges.
- The force is inversely proportional to the square of the distance between them.
- The force acts along the line joining the two charges. It is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
1.2. Mathematical Formulation (Scalar Form)
Let's translate this into an equation. Consider two point charges, $q_1$ and $q_2$, separated by a distance $r$. The magnitude of the electrostatic force ($F$) between them is given by:
$$F = k frac{|q_1 q_2|}{r^2}$$
Here's what each term represents:
- $F$: The magnitude of the electrostatic force (measured in Newtons, N).
- $q_1, q_2$: The magnitudes of the two point charges (measured in Coulombs, C). We use the absolute value $|q_1 q_2|$ because this form only gives the magnitude; the direction is determined by the signs of the charges.
- $r$: The distance between the centers of the two charges (measured in meters, m).
- $k$: This is the electrostatic constant or Coulomb's constant. It's a proportionality constant that depends on the medium in which the charges are placed.
In vacuum (or air, for practical purposes), the value of $k$ is approximately:
$$k = 9 imes 10^9 ext{ N m}^2/ ext{C}^2$$
This constant $k$ is often expressed in terms of another fundamental constant, the
permittivity of free space, denoted by $epsilon_0$ (epsilon naught).
$$k = frac{1}{4piepsilon_0}$$
The value of $epsilon_0$ is approximately $8.854 imes 10^{-12} ext{ C}^2/ ext{N m}^2$.
So, Coulomb's Law in vacuum can also be written as:
$$F = frac{1}{4piepsilon_0} frac{|q_1 q_2|}{r^2}$$
Units Check: A charge of 1 Coulomb is a very large amount of charge. Typically, in problems, you'll encounter charges in microcoulombs ($mu ext{C} = 10^{-6} ext{ C}$) or nanocoulombs ($ ext{nC} = 10^{-9} ext{ C}$).
1.3. Vector Form of Coulomb's Law
The scalar form gives us only the magnitude. To fully describe the force, we need its direction. This is where the
vector form comes in, which is crucial for JEE problems and understanding how forces combine.
Let $vec{r}_1$ and $vec{r}_2$ be the position vectors of charges $q_1$ and $q_2$, respectively.
The displacement vector from $q_1$ to $q_2$ is $vec{r}_{12} = vec{r}_2 - vec{r}_1$.
The unit vector pointing from $q_1$ to $q_2$ is $hat{r}_{12} = frac{vec{r}_{12}}{|vec{r}_{12}|}$.
The force exerted on $q_2$ by $q_1$, denoted as $vec{F}_{21}$, is:
$$vec{F}_{21} = k frac{q_1 q_2}{r^2} hat{r}_{12}$$
Substituting $hat{r}_{12} = frac{vec{r}_{12}}{|vec{r}_{12}|}$:
$$vec{F}_{21} = k frac{q_1 q_2}{|vec{r}_{12}|^3} vec{r}_{12} = k frac{q_1 q_2}{|vec{r}_2 - vec{r}_1|^3} (vec{r}_2 - vec{r}_1)$$
Important Note:
- In the vector form, we use the actual signs of $q_1$ and $q_2$.
- If $q_1 q_2$ is positive (same signs), $vec{F}_{21}$ will be in the direction of $vec{r}_{12}$, indicating repulsion.
- If $q_1 q_2$ is negative (opposite signs), $vec{F}_{21}$ will be in the opposite direction of $vec{r}_{12}$, indicating attraction.
- By Newton's Third Law, $vec{F}_{12} = -vec{F}_{21}$, meaning the forces are equal in magnitude and opposite in direction.
1.4. Medium Dependence: Dielectric Constant
When charges are placed in a medium other than vacuum, the force between them changes. The medium modifies the interaction. This effect is quantified by the
dielectric constant (or relative permittivity), denoted by $kappa$ or $epsilon_r$.
The force between two charges $q_1$ and $q_2$ separated by a distance $r$ in a medium with dielectric constant $kappa$ is:
$$F_{medium} = frac{1}{4piepsilon} frac{|q_1 q_2|}{r^2}$$
where $epsilon = epsilon_0 kappa$ is the
absolute permittivity of the medium.
Therefore, $F_{medium} = frac{1}{kappa} left( frac{1}{4piepsilon_0} frac{|q_1 q_2|}{r^2}
ight) = frac{F_{vacuum}}{kappa}$
This means that the electrostatic force in any medium is $mathbf{frac{1}{kappa}}$ times the force in vacuum. Since $kappa ge 1$ for all real materials ($kappa = 1$ for vacuum), the force between charges *always decreases* when placed in a material medium.
CBSE vs. JEE Focus:
- For CBSE, understanding the scalar form, the constant $k$, and basic calculations are key.
- For JEE, the vector form, implications of the sign of charges, the dielectric constant, and solving problems involving charges in 2D/3D coordinate systems are critically important.
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2. The Superposition Principle: Handling Multiple Charges
Coulomb's law brilliantly describes the force between *two* point charges. But what if we have three, four, or even a million charges interacting with each other? This is where the
Principle of Superposition comes to our rescue.
2.1. Statement of the Principle
The Principle of Superposition states that:
"The total electrostatic force on any one charge due to a number of other charges is the vector sum of all the individual forces exerted on that charge by all the other charges, taken one at a time. The individual forces between any two charges are unaffected by the presence of other charges."
This principle is incredibly powerful because it simplifies complex multi-charge problems into a series of two-charge problems, which we already know how to solve using Coulomb's Law.
2.2. Mathematical Formulation
Consider a system of $N$ point charges $q_1, q_2, ldots, q_N$. We want to find the net force on charge $q_1$.
According to the superposition principle, the total force $vec{F}_1$ on $q_1$ is:
$$vec{F}_1 = vec{F}_{12} + vec{F}_{13} + vec{F}_{14} + ldots + vec{F}_{1N}$$
Where $vec{F}_{1i}$ is the force on charge $q_1$ due to charge $q_i$. Each of these individual forces $vec{F}_{1i}$ is calculated using Coulomb's Law, *as if only $q_1$ and $q_i$ were present*.
2.3. Step-by-Step Application
To find the net force on a charge using the superposition principle:
- Identify the Target Charge: Determine which charge you need to find the net force on.
- Identify All Interacting Charges: List all other charges that are exerting a force on the target charge.
- Calculate Individual Forces: For each pair (target charge and one interacting charge), calculate the force vector using Coulomb's Law (vector form is generally preferred for consistency, or determine magnitude and direction separately).
- Remember to consider the signs of charges to determine if the force is attractive or repulsive.
- Draw the force vectors on the target charge.
- Resolve into Components (if needed): If forces are not collinear, resolve each force vector into its components (e.g., x and y components).
- Perform Vector Sum: Add all the x-components to get $F_x$, and all the y-components to get $F_y$.
- Find Net Force: The net force vector will be $vec{F}_{net} = F_x hat{i} + F_y hat{j}$. Its magnitude will be $|vec{F}_{net}| = sqrt{F_x^2 + F_y^2}$ and direction $ heta = an^{-1}(F_y/F_x)$.
Tip for JEE: Always draw clear diagrams showing the position of charges and the direction of individual force vectors. This visual aid will prevent errors in vector addition.
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3. Illustrative Examples
Let's solidify our understanding with some examples.
Example 1: Force between two charges in vacuum (Scalar and Vector)
Two point charges, $q_1 = +2 ext{ nC}$ and $q_2 = -4 ext{ nC}$, are placed on the x-axis. $q_1$ is at $x=0$ and $q_2$ is at $x=3 ext{ cm}$. Find the electrostatic force exerted on $q_2$ by $q_1$.
Solution:
1.
Identify charges and distance:
$q_1 = +2 imes 10^{-9} ext{ C}$
$q_2 = -4 imes 10^{-9} ext{ C}$
$r = 3 ext{ cm} = 0.03 ext{ m}$
2.
Calculate magnitude using Scalar Form:
$F = k frac{|q_1 q_2|}{r^2}$
$F = (9 imes 10^9 ext{ N m}^2/ ext{C}^2) frac{|(2 imes 10^{-9} ext{ C})(-4 imes 10^{-9} ext{ C})|}{(0.03 ext{ m})^2}$
$F = (9 imes 10^9) frac{(8 imes 10^{-18})}{0.0009}$
$F = (9 imes 10^9) (8.88 imes 10^{-15})$
$F = 7.992 imes 10^{-5} ext{ N} approx 8 imes 10^{-5} ext{ N}$
3.
Determine direction: Since $q_1$ is positive and $q_2$ is negative, the force is attractive. $q_1$ is to the left of $q_2$. So, $q_1$ pulls $q_2$ towards itself, i.e., in the negative x-direction.
4.
Vector form:
Position vector of $q_1$: $vec{r}_1 = 0 hat{i}$
Position vector of $q_2$: $vec{r}_2 = 0.03 hat{i}$
Displacement vector from $q_1$ to $q_2$: $vec{r}_{12} = vec{r}_2 - vec{r}_1 = (0.03 - 0)hat{i} = 0.03 hat{i}$
Magnitude of displacement: $|vec{r}_{12}| = 0.03 ext{ m}$
Force on $q_2$ by $q_1$: $vec{F}_{21} = k frac{q_1 q_2}{|vec{r}_{12}|^3} vec{r}_{12}$
$vec{F}_{21} = (9 imes 10^9) frac{(2 imes 10^{-9})(-4 imes 10^{-9})}{(0.03)^3} (0.03 hat{i})$
$vec{F}_{21} = (9 imes 10^9) frac{-8 imes 10^{-18}}{2.7 imes 10^{-5}} (0.03 hat{i})$
$vec{F}_{21} = (9 imes 10^9) (-2.96 imes 10^{-13}) (0.03 hat{i})$
$vec{F}_{21} = -7.992 imes 10^{-5} hat{i} ext{ N}$
The negative sign in the vector form confirms that the force is in the negative x-direction, which means $q_1$ attracts $q_2$ towards itself.
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Example 2: Superposition Principle - Three charges in a line
Three point charges $q_1 = +1 ext{ nC}$, $q_2 = -2 ext{ nC}$, and $q_3 = +3 ext{ nC}$ are placed on the x-axis at $x=0$, $x=1 ext{ m}$, and $x=2 ext{ m}$ respectively. Find the net electrostatic force on $q_2$.
Solution:
1.
Target Charge: $q_2 = -2 ext{ nC}$ at $x=1 ext{ m}$.
2.
Interacting Charges: $q_1 = +1 ext{ nC}$ at $x=0$ and $q_3 = +3 ext{ nC}$ at $x=2 ext{ m}$.
3.
Individual Forces:
*
Force on $q_2$ due to $q_1$ ($vec{F}_{21}$):
$q_1$ is positive, $q_2$ is negative. So, it's an attractive force. $q_1$ is to the left of $q_2$, so $q_2$ is pulled towards $q_1$ (i.e., in the negative x-direction).
Distance $r_{21} = 1 ext{ m}$.
Magnitude: $F_{21} = (9 imes 10^9) frac{|(1 imes 10^{-9})(-2 imes 10^{-9})|}{(1)^2} = 18 imes 10^{-9} ext{ N}$
So, $vec{F}_{21} = -18 imes 10^{-9} hat{i} ext{ N}$
*
Force on $q_2$ due to $q_3$ ($vec{F}_{23}$):
$q_3$ is positive, $q_2$ is negative. So, it's an attractive force. $q_3$ is to the right of $q_2$, so $q_2$ is pulled towards $q_3$ (i.e., in the positive x-direction).
Distance $r_{23} = 1 ext{ m}$.
Magnitude: $F_{23} = (9 imes 10^9) frac{|(-2 imes 10^{-9})(+3 imes 10^{-9})|}{(1)^2} = 54 imes 10^{-9} ext{ N}$
So, $vec{F}_{23} = +54 imes 10^{-9} hat{i} ext{ N}$
4.
Vector Sum:
The net force on $q_2$ is $vec{F}_{net} = vec{F}_{21} + vec{F}_{23}$
$vec{F}_{net} = (-18 imes 10^{-9} hat{i}) + (54 imes 10^{-9} hat{i})$
$vec{F}_{net} = (54 - 18) imes 10^{-9} hat{i}$
$vec{F}_{net} = +36 imes 10^{-9} hat{i} ext{ N}$
The net force on $q_2$ is $36 ext{ nN}$ in the positive x-direction.
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Example 3: Superposition Principle - Three charges in a triangle (JEE Advanced level)
Three charges $q_A = +Q$, $q_B = +Q$, and $q_C = -Q$ are placed at the vertices A, B, and C respectively, of an equilateral triangle of side length $a$. Find the net electrostatic force on charge $q_C$.
Solution:
1.
Target Charge: $q_C = -Q$.
2.
Interacting Charges: $q_A = +Q$ and $q_B = +Q$.
Let's place the triangle in a coordinate system. Let C be at the origin $(0,0)$.
Then B would be at $(a,0)$.
And A would be at $(a/2, asqrt{3}/2)$.
3.
Individual Forces on $q_C$:
*
Force on $q_C$ due to $q_A$ ($vec{F}_{CA}$):
$q_A = +Q$, $q_C = -Q$. The force is attractive. So $vec{F}_{CA}$ points from C towards A.
Magnitude: $F_{CA} = k frac{|(Q)(-Q)|}{a^2} = k frac{Q^2}{a^2}$
The vector direction of $vec{F}_{CA}$ is along the line CA. The angle that CA makes with the positive x-axis is $60^circ$ (since it's an equilateral triangle and C is at origin, A is up-left).
So, $vec{F}_{CA} = F_{CA} (cos 60^circ hat{i} + sin 60^circ hat{j})$
$vec{F}_{CA} = frac{kQ^2}{a^2} (frac{1}{2} hat{i} + frac{sqrt{3}}{2} hat{j})$
*
Force on $q_C$ due to $q_B$ ($vec{F}_{CB}$):
$q_B = +Q$, $q_C = -Q$. The force is attractive. So $vec{F}_{CB}$ points from C towards B.
Magnitude: $F_{CB} = k frac{|(Q)(-Q)|}{a^2} = k frac{Q^2}{a^2}$
The vector direction of $vec{F}_{CB}$ is along the positive x-axis (from C(0,0) to B(a,0)).
So, $vec{F}_{CB} = F_{CB} (1 hat{i} + 0 hat{j}) = frac{kQ^2}{a^2} hat{i}$
4.
Vector Sum:
$vec{F}_{net} = vec{F}_{CA} + vec{F}_{CB}$
$vec{F}_{net} = frac{kQ^2}{a^2} (frac{1}{2} hat{i} + frac{sqrt{3}}{2} hat{j}) + frac{kQ^2}{a^2} hat{i}$
$vec{F}_{net} = frac{kQ^2}{a^2} left[ left(frac{1}{2} + 1
ight) hat{i} + frac{sqrt{3}}{2} hat{j}
ight]$
$vec{F}_{net} = frac{kQ^2}{a^2} left[ frac{3}{2} hat{i} + frac{sqrt{3}}{2} hat{j}
ight]$
The magnitude of the net force:
$|vec{F}_{net}| = frac{kQ^2}{a^2} sqrt{left(frac{3}{2}
ight)^2 + left(frac{sqrt{3}}{2}
ight)^2}$
$|vec{F}_{net}| = frac{kQ^2}{a^2} sqrt{frac{9}{4} + frac{3}{4}}$
$|vec{F}_{net}| = frac{kQ^2}{a^2} sqrt{frac{12}{4}} = frac{kQ^2}{a^2} sqrt{3}$
This detailed breakdown of Coulomb's Law and the Superposition Principle should equip you with the fundamental tools to tackle a wide range of problems in electrostatics. Remember, practice is key, especially with the vector aspect of these laws! Keep working on varied problems, and you'll master these concepts in no time.