The Superposition Principle

1. Statement of the Principle / Concept Overview

The Superposition Principle states that:

The net electric force (or electric field) acting on a charge due to a group of charges is the vector sum of the individual forces (or fields) exerted by each charge independently.

It assumes that:

• Charges do not alter each other’s fields.
• Electric field contributions add vectorially, not algebraically.

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2. Clear Explanation and Mathematical Derivation

Consider charges [q_1, q_2, q_3, \dots, q_n] producing electric fields at a point P.

Let:

• [\vec{E}_1] = electric field at P due to [q_1]
• [\vec{E}_2] = electric field at P due to [q_2]
• …
• [\vec{E}_n] due to [q_n]

 

Then the resultant field is:

[\vec{E}_{\text{net}}] [= \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \dots + \vec{E}_n]

Since

[\vec{E}_i] [= k\dfrac{q_i}{r_i^2}\hat{r}_i,]

we have:

[\vec{E}{\text{net}}] [= k\sum{i=1}^{n}\dfrac{q_i}{r_i^2}\hat{r}_i]

Similarly, for force on a test charge [q_0]:

[\vec{F}_{\text{net}}] [= \vec{F}_1 + \vec{F}2 + \dots] [= q_0\vec{E}{\text{net}}]

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3. Dimensions and Units

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4. Key Features

• Electric fields add vectorially (direction matters).
• Works for any number of charges.
• Holds for static and continuous charge distributions.
• Results remain valid even if charges are very close.
• Fundamental for solving electric field problems in:
  • dipoles
  • continuous charge distributions
  • electric field mapping
  • Coulomb force in multi-charge systems

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5. Important Formulas to Remember

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6. Conceptual Questions With Solutions

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7. FAQ / Common Misconceptions

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Practice Questions (With Step-by-Step Solutions)

Practice Question 1

Two point charges [q_1] [= +2.0 \times 10^{-6}\ \text{C}] (at [x=0]) and [q_2] [= +2.0 \times 10^{-6}\ \text{C}] (at [x=0.30\ \text{m}]) lie on the x-axis.
Find the net electric field at the point [x=0.60\ \text{m}] (on the x-axis). Give magnitude and direction.

Solution (step-by-step):

1. Distance from [q_1] to the point: [r_1 = 0.60\ \text{m}].
   Distance from [q_2] to the point: [r_2 = 0.60 – 0.30 = 0.30\ \text{m}].
2. Field due to [q_1] at the point (points away from +ve charge ⇒ to the +x direction):
   [E_1] [= k \dfrac{q_1}{r_1^2}] [= 9\times 10^{9}\dfrac{2.0\times 10^{-6}}{(0.60)^2}]
   Compute: [9\times10^{9}\times 2.0\times10^{-6}] [= 1.8\times10^{4}].
   [(0.60)^2 = 0.36].
   [E_1 = \dfrac{1.8\times10^{4}}{0.36}] [= 5.0\times10^{4}\ \text{N/C}]
3. Field due to [q_2] at the point (also points away, i.e. +x):
   [E_2] [= 9\times10^{9}\dfrac{2.0\times10^{-6}}{(0.30)^2}]
   [(0.30)^2 = 0.09.]
   [E_2 = \dfrac{1.8\times10^{4}}{0.09}] [= 2.0\times10^{5}\ \text{N/C}]
4. Net electric field (same direction, add algebraically):
   [E_{\text{net}}] [= E_1 + E_2 = 5.0\times10^{4} + 2.0\times10^{5}] [= 2.50\times10^{5}\ \text{N/C}]

Answer: [2.50\times10^{5}\ \text{N/C}] toward the +x direction.

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Practice Question 2

Two point charges [q_1] [= +4.0\times10^{-6}\ \text{C}] at [x=-0.10\ \text{m}] and [q_2 = -4.0\times10^{-6}\ \text{C}] at [x=+0.10\ \text{m}]. Find the net electric field at the origin [(x=0)].

Solution:

1. Distance from each charge to origin: [r = 0.10\ \text{m}].
2. Magnitude of field from either charge:
   [E = k \dfrac{4.0\times10^{-6}}{(0.10)^2}] [= 9\times10^{9}\dfrac{4.0\times10^{-6}}{0.01}]
   Numerator: [9\times10^{9}\times 4.0\times10^{-6}] [= 3.6\times10^{4}].
   Divide: [\dfrac{3.6\times10^{4}}{0.01}] [= 3.6\times10^{6}\ \text{N/C}.]
3. Directions at origin:
   • Field due to [q_1=+4\mu\text{C}] (at [x=-0.10]) points away from q1 → towards +x. So [E_1] [= +3.6\times10^{6}\ \text{N/C}].
   • Field due to [q_2=-4\mu\text{C}] (at [x=+0.10]) points toward q2 → also towards +x. So [E_2] [= +3.6\times10^{6}\ \text{N/C}].
4. Net field:
   [E_{\text{net}}] [= E_1 + E_2] [= 3.6\times10^{6} + 3.6\times10^{6}] [= 7.2\times10^{6}\ \text{N/C}]

Answer: [7.2\times10^{6}\ \text{N/C}] in the +x direction.

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Practice Question 3

A test charge [q_0] [= +1.0\times10^{-6}\ \text{C}] is placed at the midpoint between [q_1] [= +3.0\times10^{-6}\ \text{C}] (at [x=0]) and [q_2] [= -3.0\times10^{-6}\ \text{C}] (at [x=0.20\ \text{m}]). Find the force on [q_0].

Solution:

1. Midpoint location: [x=0.10\ \text{m}]. Distances from each charge: [r_1 = r_2 = 0.10\ \text{m}].
2. Field at midpoint due to [q_1] (positive):
   [E_1 = k\dfrac{3.0\times10^{-6}}{(0.10)^2}] [= 9\times10^{9}\dfrac{3.0\times10^{-6}}{0.01}]
   [E_1 = \dfrac{2.7\times10^{4}}{0.01}] [= 2.7\times10^{6}\ \text{N/C}]
   Direction: away from q1 ⇒ +x.
3. Field at midpoint due to [q_2] (negative): field points toward q2. Since q2 is at +x, field at midpoint is also +x with same magnitude: [E_2 = 2.7\times10^{6}\ \text{N/C}].
4. Net field:
   [E_{\text{net}}] [= E_1 + E_2] [= 2.7\times10^{6} + 2.7\times10^{6}] [= 5.4\times10^{6}\ \text{N/C}]
5. Force on test charge:
   [F = q_0 E_{\text{net}}] [= 1.0\times10^{-6}\times 5.4\times10^{6}] [= 5.4\ \text{N}]

Direction: +x.

Answer: [F = 5.4\ \text{N}] to the +x direction.

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Practice Question 4 (Potential — scalar superposition)

Two equal charges [q] [= +2.0\times10^{-6}\ \text{C}] are placed at [x=-0.20\ \text{m}] and [x=+0.20\ \text{m}]. Find the electric potential at the origin.

Solution:

1. Potential due to a point charge at distance [r] is: [V = k\dfrac{q}{r}]. Potentials add algebraically (scalar).
2. Distance from each charge to origin: [r=0.20\ \text{m}].
3. Potential from one charge:
   [V_1] [= 9\times10^{9}\dfrac{2.0\times10^{-6}}{0.20}] [= 9\times10^{9}\times \dfrac{2.0\times10^{-6}}{0.2}]
   Compute numerator: [9\times10^{9}\times 2.0\times10^{-6}] [= 1.8\times10^{4}.]
   Divide by 0.2: [\dfrac{1.8\times10^{4}}{0.2}] [= 9.0\times10^{4}\ \text{V}.]
4. Since both charges identical and potential is scalar:
   [V_{\text{total}}] [= V_1 + V_2] [= 9.0\times10^{4} + 9.0\times10^{4}] [= 1.8\times10^{5}\ \text{V}]

Answer: [1.8\times10^{5}\ \text{V}] at the origin.

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Practice Question 5 (2-D vector addition)

Charge [q_1] [= +6.0\times10^{-6}\ \text{C}] is at the origin [(0,0)]. Charge [q_2] [= -2.0\times10^{-6}\ \text{C}] is at [(0.30\ \text{m},,0)]. Find the net electric field at point [P(0,;0.40\ \text{m})]. Give the vector components, magnitude and direction (angle w.r.t +x axis).

Solution:

1. Position vectors from sources to P:
   • From [q_1]: [(\vec{r}_1 = (0,;0.40))], so [r_1 = 0.40\ \text{m}.]
   • From [q_2]: [(\vec{r}_2 = (-0.30,;0.40))] (point P minus q2 position), so [r_2] [= \sqrt{(-0.30)^2 + (0.40)^2}] [= \sqrt{0.09 + 0.16}] [= \sqrt{0.25} = 0.50\ \text{m}.]
2. Magnitude of field from [q_1]:
   [E_1] [= k \dfrac{6.0\times10^{-6}}{(0.40)^2}] [= 9\times10^{9}\dfrac{6.0\times10^{-6}}{0.16}]
   Numerator: [9\times10^{9}\times 6.0\times10^{-6}] [= 5.4\times10^{4}.]
   [E_1] [= \dfrac{5.4\times10^{4}}{0.16}] [= 3.375\times10^{5}\ \text{N/C}]
   Direction: away from +ve charge → along +y (i.e., vector [(0,;3.375\times10^{5})] N/C).
3. Magnitude of field from [q_2] (charge is negative → field points toward q2, i.e. along vector from P to q2 which is ((+0.30,,-0.40)) direction):
   [E_2] [= k \dfrac{2.0\times10^{-6}}{(0.50)^2}] [= 9\times10^{9}\dfrac{2.0\times10^{-6}}{0.25}]
   Numerator: [9\times10^{9}\times 2.0\times10^{-6}] [= 1.8\times10^{4}.]
   [E_2] [= \dfrac{1.8\times10^{4}}{0.25}] [= 7.2\times10^{4}\ \text{N/C}]Unit vector from P toward q2:
   [\hat{u}_2] [= \dfrac{(0.30,,-0.40)}{0.50}] [= (0.6,,-0.8).]So vector field from [q_2]:
   [\vec{E}_2] [= 7.2\times10^{4}\times(0.6,,-0.8)] [= (4.32\times10^{4},,-5.76\times10^{4})\ \text{N/C}.]
4. Add vectors:
   [\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2] [= \big(4.32\times10^{4},;3.375\times10^{5} – 5.76\times10^{4}\big)]
   Compute y-component: [3.375\times10^{5} – 5.76\times10^{4} = 2.799\times10^{5}.]So
   [\vec{E}_{\text{net}}] [= (4.32\times10^{4},;2.799\times10^{5})\ \text{N/C}.]
5. Magnitude:
   [|\vec{E}|] [= \sqrt{(4.32\times10^{4})^2 + (2.799\times10^{5})^2} \approx 2.83214\times10^{5}\ \text{N/C}]
6. Direction (angle above +x axis):
   [\theta] [= \tan^{-1}!\Big(\dfrac{2.799\times10^{5}}{4.32\times10^{4}}\Big) \approx 81.23^\circ.]

Answer:
[\vec{E}_{\text{net}}] [= (4.32\times10^{4}\ \hat{i} + 2.799\times10^{5}\ \hat{j})\ \text{N/C},]
magnitude [2.83\times10^{5}\ \text{N/C}], at [81.2^\circ] above +x.

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Practice Question 6 (Combine many contributions)

Four equal charges, each [+1.0\times10^{-6}\ \text{C}], are placed at the corners of a square of side [0.10\ \text{m}]. Find the electric field at the center of the square.

Solution (sketch with steps):

1. By symmetry, fields from opposite corners have same magnitude and directions that combine. Let center be O. Distance from a corner to center: [r] [= \dfrac{\sqrt{2}}{2}\times 0.10] [= 0.07071\ \text{m}.]
2. Field magnitude from one charge at center:
   [E_{\text{one}}] [= k\dfrac{1.0\times10^{-6}}{r^2}] [= 9\times10^{9}\dfrac{1.0\times10^{-6}}{(0.07071)^2}].
   Compute [r^2] [= (0.07071)^2 \approx 0.005].
   Numerator: [9\times10^{9}\times1.0\times10^{-6}] [= 9.0\times10^{3}.]
   [E_{\text{one}} \approx \dfrac{9.0\times10^{3}}{0.005}] [= 1.8\times10^{6}\ \text{N/C} \ (\text{approx.})]
3. Direction: each field points radially away from its corner (since charges are +ve). Pairwise, contributions from diagonally opposite corners cancel in perpendicular directions leaving zero net (because four identical symmetric vectors sum to zero).
4. By symmetry the net electric field at the center due to four equal charges at corners of a square is zero (all radial components cancel).

Answer: [ \vec{E}_{\text{center}}] [= \vec{0}.]

(Note: step 3 shows why — symmetry causes complete cancellation.)