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.
Key equation
Fnet=F1+F2+F3+⋯=∑iFi
\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
Key equation
Fij=k|qiqj|rij2
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.
Key equation
Fx=Fcosθ,\qquad Fy=Fsinθ
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)
Key equation
Fnet,x=∑iFix,\qquad Fnet,y=∑iFiy
Key equation
Fnet=Fnet,xi+Fnet,yj
\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.
Key equation
q1=q3=5.00μC=5.00×10-6C
Key equation
q2=-2.00μC=-2.00×10-6C,\qquad a=0.100m
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
Key equation
F32=14πε0|q2q3|a2
Key equation
F32=9×109(2×10-6)(5×10-6)(0.100)2=8.99N
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.
Key equation
F31=14πε0|q1q3|(√2a)2=14πε0|q1q3|2a2
Key equation
F31=9×109(5×10-6)(5×10-6)2(0.100)2=11.2N
Key equation
F31x=F31cos45°=7.94N
Key equation
F31y=F31sin45°=7.94N
Resultant force on q₃
Key equation
F3x=F31x+F32x=7.94+(-8.99)=-1.04N
Key equation
F3y=F31y+F32y=7.94+0=7.94N
Key equation
F3=(-1.04i+7.94j)N
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
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Exam preparation
Exam Focus
Important points
Vector-sum statement.
Independent pairwise forces.
Attraction and repulsion directions.
Resolution into signed components.
Textbook Example 10.1 result.
Common mistakes
Adding force magnitudes while ignoring their directions.
Choosing attraction or repulsion incorrectly from the charge signs.
Using the side length a instead of the diagonal distance √2a.
Dropping negative component signs during vector addition.
Substituting microcoulombs without converting them to coulombs.
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.
Superposition of Electrostatic Forces
Sticky Notes — Quick Revision
1.
Key Point 1
Pairwise forces are calculated separately as if the other source charges were absent.
2.
Key Point 2
The net force is the vector sum F⃗net=ΣF⃗i.
3.
Key Point 3
For two-dimensional arrangements, resolve every force into x- and y-components.
4.
Key Point 4
In textbook Example 10.1, the resultant on q₃ is F⃗₃=(−1.04î+7.94ĵ) N.
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8 concept questions
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Explanation after each answer
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?
