Electric Field Strength from point charges

In summary, when a particle with charge q is placed at point P, the electric field at point P is composed of two components, each with a magnitude of kQ/(L^2)^(1/2).
  • #1
pingpong240
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0
Question

Charges at three corners of a square are shown in Figure P26.34, in which n = 9. To answer the following questions, use the letters k, Q, q, L, and m.
IMAGE IS LOCATED HERE: http://www.webassign.net/knight/p26-34.gif
a) Write the electric field at point P in component form.
b)A particle with positive charge q and mass m is placed at point P and released. What is the initial magnitude of its acceleration?
Attempt at a Solution

Well, I was a little upset about this problem because I wonder if I'm getting the typing syntax correct. Anyways, here's what I came up with for an answer for the x coordinate of a(which is wrong):
-k(Q/(L^2))+kcos(45)((9Q)/(2L^2)^(1/2))

k is of course Coulomb's constant and Q is the charge, with L as the distance

Any ideas on what I am doing wrong here?
 
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  • #2
The electric field obeys superposition. This means that, to find the field from multiple charges, you just add up the field from each individual charge.

It looks like you're on the right track, but the *square* of the distance from nQ to P is 2L², not the square root of that.
 
  • #3
yeah that's what i have i think, i double checked the format and what i have is equivalent to (L^2+L^2)^(1/2) or 2(L)^(1/2)

that's what you are talking about right?
i know the other charge doesn't get involved in the x coordinate...so what might i be doing wrong?
 
  • #4
You calculated kcos(45)((9Q)/(2L^2)^(1/2)) as the contribution to E from nQ.

The number in the denominator, (2L²)^(1/2), is the distance from nQ to P. But your formula for the field magnitude, E(r) = kq/r², puts the square of the distance in the denominator. You are using the distance itself, not the square of the distance.
 
  • #5
ah i see now. thank you very much i got it.
 

1. What is electric field strength from point charges?

Electric field strength from point charges is the measure of the force per unit charge experienced by a test charge placed in the electric field created by a point charge. It is a vector quantity and is expressed in units of newtons per coulomb (N/C).

2. How is electric field strength from point charges calculated?

The electric field strength from point charges can be calculated using the formula E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the magnitude of the point charge, and r is the distance between the point charge and the test charge.

3. What factors affect the electric field strength from point charges?

The electric field strength from point charges is affected by the magnitude of the point charge, the distance between the point charge and the test charge, and the medium in which the charges are located. The electric field strength also follows the inverse square law, which means that it decreases as the distance from the point charge increases.

4. How does the direction of the electric field strength from point charges relate to the direction of the electric field?

The electric field strength from point charges is a vector quantity, meaning it has both magnitude and direction. The direction of the electric field strength is always in the same direction as the electric field, which is the direction a positive test charge would move in the presence of the electric field.

5. What are some real-world applications of electric field strength from point charges?

Electric field strength from point charges is used in a variety of real-world applications, including electrical power transmission, capacitors, and electrostatic precipitators. It is also a fundamental concept in understanding the behavior of charged particles in electric and magnetic fields, which is essential in fields such as electronics, particle physics, and astronomy.

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