Calculating Electric Field at Point D: Two Equal Charges

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Homework Help Overview

The problem involves calculating the electric field at a specific point (D) due to two equal charges located at the corners of a square. The charges are given as Q = +0.80 nC, and the side length of the square is x = 1.2 m.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field from each charge separately and consider the vector nature of the fields. There are attempts to apply the Pythagorean theorem to find the resultant electric field, and questions arise regarding the distances used in calculations.

Discussion Status

The discussion includes various attempts to calculate the electric field, with some participants suggesting different methods and questioning the correctness of their calculations. There is no explicit consensus on the final answer, but several approaches have been explored.

Contextual Notes

Participants note the importance of using the correct distance in calculations and the need to consider the vector nature of electric fields when combining them. Some calculations yield different results, prompting further discussion on the methodology.

Bob Loblaw
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Homework Statement


Two equal charges (Q = +0.80 nC) are situated at the diagonal corners A and B of a square of side x = 1.2 m as shown in the diagram. What is the magnitude of the electric field at point D?
p16-38.gif


Homework Equations



E=kQ/r^2

The Attempt at a Solution



First I converted 0.80 nC into C and got 8.0e-10 C.

Using pythagorean's theorum I found 1.7 m as the distance between the two charges.

I then used Coulomb's law as follows:

9e9*8e-10/1.7^2 which gave me 2.5.

Since we want point D, I took the sin(45)2.5 + cos(45)2.5 and came up with 3.52 which was wrong.

Any ideas where I goofed?
 
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The total field at D can just be considered as the field from particle A + the field from particle B. So the total equation is just E = 2 * k * Q / x^2
 
sarujin said:
The total field at D can just be considered as the field from particle A + the field from particle B. So the total equation is just E = 2 * k * Q / x^2

Thanks for the reply. I took 2(9e9*8e-10)/1.7^2 and came up with 4.98 which did not seem to work.
 
how about

square root of (2.5^2 + 2.5^2)?
 
Leong said:
how about

square root of (2.5^2 + 2.5^2)?

Tried that earlier and it did not seem to work.
 
how about

square root of (5^2 + 5^2) = 7.07?
 
Leong said:
how about

square root of (5^2 + 5^2) = 7.07?

That answer was accepted. Thank you. If you have a minute, would you mind telling me how you came up with that?
 
1. electric field at D due to A is acting downward: magnitude 5
2. electric field at D due to B is acting to the left:magnitude 5
3. these two electric fields are vectors and are perpendicular to each other
4. if you are only interested in finding the magnitude of the resultant field,you can just apply pythagorean theorem.
 
Thank you, Leong.
 
  • #10
make sure you use 1.2 m instead of 1.7 m because we find the field separately and individually; one by one and then only we find the overall/ resultant field.
 
  • #11
glad to help and see that you really appreciate that...
 

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