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Superposition of Electrostatic Forces
The superposition principle states that the net electrostatic force on a charge equals the vector sum of the separate Coulomb forces exerted by all other charges. Each pairwise force is calculated independently, resolved into components when necessary, and then added component by component.
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The main idea
When several charges surround one charge, each produces a push or pull. Calculate every force separately, keep its direction, and then add all the force vectors.
- \vec F_{net}
- resultant force on the selected charge (N)
- \vec F_i
- individual force from source charge i (N)
Simple method
- Choose the charge on which the net force is required.
- Draw attraction or repulsion from every other charge.
- Use Coulomb's law for each force magnitude.
- Resolve sloping forces into components.
- Add x-components and y-components separately.
Recommended for Class 11
Statement of the superposition principle
If several charges are present, a selected charge experiences a separate electrostatic force due to every other charge. The resultant force is found by adding these individual forces as vectors.
- \vec F_i
- pairwise force exerted by the ith source charge (N)
Each pairwise force is calculated one at a time using Coulomb's law. The presence of additional charges does not change the mathematical form of a particular pair interaction.
Pairwise Coulomb forces
- F_{ij}
- magnitude of the force between charges i and j (N)
- r_{ij}
- separation between the two charges (m)
| Charge signs | Direction of force on the selected charge |
|---|---|
| Same sign | Away from the source charge because the interaction is repulsive |
| Opposite signs | Toward the source charge because the interaction is attractive |
Vector-component method
Forces along different directions cannot be added as ordinary scalar numbers. Resolve a force F making angle θ with the positive x-axis into rectangular components.
- F_x
- x-component of force (N)
- F_y
- y-component of force (N)
- \theta
- angle measured from the positive x-axis (degree or radian)
- \hat i
- unit vector along positive x (none)
- \hat j
- unit vector along positive y (none)
A component pointing left or downward is negative; a component pointing right or upward is positive.
Textbook Example 10.1: arrangement
Three point charges occupy the corners of a right triangle. The given values are q₁=q₃=5.00 μC, q₂=−2.00 μC, and a=0.100 m. The resultant force on q₃ is required.
Charge q₃ experiences two forces. Because q₂ and q₃ have opposite signs, F⃗₃₂ is attractive and points horizontally toward q₂. Because q₁ and q₃ are both positive, F⃗₃₁ is repulsive and points at 45° above the positive x-axis.
Force on q₃ due to q₂
The force points to the left, so its components are F₃₂x=−8.99 N and F₃₂y=0.
Force on q₃ due to q₁
The diagonal separation between q₁ and q₃ is √2a, so the squared separation is 2a².
Resultant force on q₃
General solution procedure
- Select the charge on which the net force is required.
- Draw a separate force arrow from every other charge.
- Label each force as attractive or repulsive.
- Calculate each magnitude with Coulomb's law using SI units.
- Resolve every oblique force into signed components.
- Add all x-components and all y-components.
- Write the final answer in vector form and interpret its direction.
| Common situation | Efficient approach |
|---|---|
| All forces along one line | Choose one positive direction and add signed forces algebraically |
| Forces at right angles | Add x- and y-components separately |
| Symmetric arrangement | Look for equal components that cancel before calculating |
| General two-dimensional arrangement | Use trigonometric components for every force |
Exam Focus
Exam Focus
Important points
- Vector-sum statement.
- Independent pairwise forces.
- Attraction and repulsion directions.
- Resolution into signed components.
- Textbook Example 10.1 result.
Important formulas
- Fnet=∑i Fi
- Fx=Fcosθ
- Fy=Fsinθ
- F3=(-1.04 i+7.94 j)N
Common exam traps
- Use √2a for the diagonal separation in Example 10.1.
- A leftward component is negative when +x points right.
Long question structure
Statement → force diagram → pairwise Coulomb magnitudes → components → signed sums → final vector and direction
Common Mistakes
Common Mistakes
- Mistake
Adding force magnitudes while ignoring their directions.
- Mistake
Choosing attraction or repulsion incorrectly from the charge signs.
- Mistake
Using the side length a instead of the diagonal distance √2a.
- Mistake
Dropping negative component signs during vector addition.
- Mistake
Substituting microcoulombs without converting them to coulombs.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Each surrounding charge exerts its own Coulomb force on the selected charge.
Pairwise forces are calculated separately as if the other source charges were absent.
The net force is the vector sum F⃗net=ΣF⃗i.
Force directions must be decided from attraction or repulsion before addition.
The net force on a charge equals the vector sum of the separate Coulomb forces exerted by every other charge.
Fnet=∑i Fi, Fx=Fcosθ, Fy=Fsinθ
- Definition
- The net force on a charge equals the vector sum of the separate Coulomb forces exerted by every other charge.
- Formula
- \vec F{net}=∑i\vec Fi,\quad Fx=F\cos\theta,\quad F_y=F\sin\theta
- Example
- For textbook Example 10.1, F⃗₃=(−1.04î+7.94ĵ) N.
- Critical warning
- Resolve forces into signed components before adding them.
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Exam Practice
Important Question
Q1. State and explain the superposition principle of electrostatic force.
Method
Q2. How is the net electrostatic force found in two dimensions?
Numerical
Q3. Reproduce the component result of textbook Example 10.1 for q₃.
Reasoning
Q4. Why must electrostatic forces be added as vectors?
Concept
Q5. Does the presence of a third charge change the pairwise Coulomb force formula between two selected charges?