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Physics Lesson 14.3.4 - The Superposition Principle

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Welcome to our Physics lesson on The Superposition Principle, this is the fourth lesson of our suite of physics lessons covering the topic of Electric Field, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

The Superposition Principle

If in a given region of space there is more than one charge that produce their own electric fields E1, E2, + En, the net electric field in that region is the sum of all individual electric field vectors, i.e.

Enet = E1 + E2 + ... + En

Example 2

Two point charges Q1 and Q2 where Q1 = + 20 μC and Q2 = - 30 μC are placed as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Field
  1. What is the magnitude of net electric field at the point P?
  2. What is the angle formed by the vector of net electric field at P to the horizontal direction?

Solution 2

a) First, we find the distance from Q1 to P. Using the Pythagorean Theorem, we have

r(Q1P) = √r(Q1 Q2 )2 - r(Q2 P)2
= √152 - 92
= √225 - 81
= √144
= 12 cm

Now, we find the individual electric fields E1P and E2P where E1P vector is only horizontal (due right, as the electric field lines start from Q2 and extend to infinity) and E2P is only vertical (upward, as the electric field lines start from infinity and end to Q2 because the charge Q2 is negative).

We have:

E1P = 1/4πϵ0 × Q1/r1P2
= 9 × 109 × 20 × 10-6/0.122
= 180 × 10-6/0.0144
= 1.8 × 10-4/1.44 × 10-2
= 1.25 × 10-2 N/C

Also,

E2P = 1/4πϵ0 × Q2/r2P2
= 9 × 109 × -30 × 10-6/0.092
= -270 × 10-6/0.0081
= -27 × 10-5/8.1 × 10-3
= -3.33 × 10-2 N/C

The magnitude of the resultant electric field at the point P therefore is

EP = √E1P2 + E2P2
= √(1.25 × 10-2 )2 + (-3.33 × 10-2 )2
= √1.625 × 10-4 + 11.111 × 10-4
= √12.736 × 10-4
= 3.569 × 10-2 N/C

The vector of resultant electric field at the point P is shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Electric Field

b) The angle θ formed by the vector of resultant electric field to the horizontal direction is obtained by calculating its tangent. Thus, we have

tan θ = |E2P|/|E1P|
= |-3.33| × 10-2/|1.25| × 10-2
= 3.33/1.25
= 2.664

Therefore, the angle θ is

θ = arctan 2.664
= 69.40

You have reached the end of Physics lesson 14.3.4 The Superposition Principle. There are 6 lessons in this physics tutorial covering Electric Field, you can access all the lessons from this tutorial below.

More Electric Field Lessons and Learning Resources

Electrostatics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
14.3Electric Field
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
14.3.1What Is Electric Field? Similarities between Gravitational and Electric Field
14.3.2Electric Field Lines
14.3.3Electrostatic Constants
14.3.4The Superposition Principle
14.3.5Electric Field on a Charged Sphere
14.3.6Conductors in an Electric Field

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  4. Electrostatics Practice Questions: Electric Field. Test and improve your knowledge of Electric Field with example questins and answers
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  6. Continuing learning electrostatics - read our next physics tutorial: Electric Potential Energy

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