Calculating Electric Field at Point D: Two Equal Charges

In summary, an electric field of 5 volts/meter is present at point D due to the charges of 0.80 nC situated at the diagonal corners of a square of side 1.2 m.
  • #1
Bob Loblaw
69
0

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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  • #2
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
 
  • #3
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.
 
  • #4
how about

square root of (2.5^2 + 2.5^2)?
 
  • #5
Leong said:
how about

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

Tried that earlier and it did not seem to work.
 
  • #6
how about

square root of (5^2 + 5^2) = 7.07?
 
  • #7
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?
 
  • #8
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.
 
  • #9
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...
 

1. What is the formula for calculating electric field at point D?

The formula for calculating the electric field at point D is E = kQ/r^2, where E is the electric field strength, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the magnitude of the charges, and r is the distance between the charges and point D.

2. How do I determine the direction of the electric field at point D?

The direction of the electric field at point D is determined by the direction of the force that a positive test charge would experience if placed at that point. The direction of the electric field is always away from a positive charge and towards a negative charge.

3. Can I use this formula for multiple charges at point D?

Yes, this formula can be used for multiple charges at point D by simply adding the electric fields from each individual charge using vector addition. Remember to consider the direction of each individual electric field when using vector addition.

4. How does the distance between the charges and point D affect the electric field?

The electric field at point D is inversely proportional to the square of the distance between the charges and point D. This means that as the distance increases, the electric field decreases. This relationship is described by the inverse square law.

5. What are the units for electric field at point D?

The units for electric field at point D are newtons per coulomb (N/C) or volts per meter (V/m). These units represent the amount of force experienced by a positive test charge per unit charge.

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