Conceptual: two positive point charges value of E

In summary, the conversation discusses the effects of two charges, Q1 and Q2, on the electric fields at point P. The principle of superposition applies in classical electrodynamics, but in quantum electrodynamics, fields become very strong and the Schwinger limit applies. The two charges exert destructive interference, but the waves do not disappear and combine after passing each other, diminishing as 1/r^2. The question asks why Q1's electric field is not affected by the repulsion of the two charges, and the answer lies in the principle of superposition.
  • #1
origamipro
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Homework Statement



Both Q1 = Q2 = +5E-6 and are harnessed to screen and in a vacuum.----Q1---A----Q2------B--------P this is x axis, A and B are distances

Homework Equations



kQ/r^2

The Attempt at a Solution



So the E field at point P is KQ1/(A+B)^2 + KQ2/(B)^2
My question is Why isn't Q1 E field affected by the repulsion of the two charges. IE if photons carry the force of the charged particle, why arent the photons affected by the presence of Q2 emitting photons 360 degrees. So to the left of Q2, Q2 sends its signed wave to the left while a same signed wave is moving to the right from Q1. Shouldnt they interact, that is how they would repel if not anchored to the screen.
 
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  • #2
In classical electrodynamics, electric fields are governed by the principle of superposition, i.e., the total field at any point in space is the sum of the fields due to individual sources. In quantum electrodynamics, the principle of superposition holds until fields become very strong. This is known as the Schwinger limit.
 
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  • #3
Ok i see it now. The two charges exert destructive interference, but the waves don't disappear. Once Q1 wave passes Q2 it combines and it diminishes as 1/r^2.
 
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Related to Conceptual: two positive point charges value of E

1. What is the formula for calculating the value of electric field (E) between two positive point charges?

The formula for calculating the value of electric field (E) between two positive point charges is E = kQ/r^2, where k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), Q is the magnitude of the charges, and r is the distance between the charges.

2. How does the distance between two positive point charges affect the value of electric field (E)?

The value of electric field (E) between two positive point charges is inversely proportional to the square of the distance between the charges. This means that as the distance between the charges increases, the value of electric field decreases and vice versa.

3. Can the value of electric field (E) between two positive point charges be negative?

No, the value of electric field (E) between two positive point charges cannot be negative. Electric field is a vector quantity and its direction is determined by the direction of the force on a positive test charge placed in the field. Since like charges repel each other, the direction of the electric field between two positive charges is always away from the charges.

4. How is the value of electric field (E) affected if the magnitude of the charges is doubled?

If the magnitude of the charges is doubled, the value of electric field (E) will also double. This is because the electric field is directly proportional to the magnitude of the charges. Therefore, increasing the charge will result in a stronger electric field.

5. Can the value of electric field (E) between two positive point charges be zero?

Yes, the value of electric field (E) between two positive point charges can be zero. This occurs when the charges are equal in magnitude and are placed at equal distances from each other. In this case, the electric forces will cancel each other out and there will be no net electric field between the charges.

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