I need some help on how to calculate electric field vectors at a point called P

In summary: Thanks for looking.In summary, the student is studying for a physics midterm and got stuck on one of the problems. They solved it themselves and found the angles for the x and y components of the electric field at point P.
  • #1
tastybrownies
8
0
Hello everyone and I hope you are enjoying your weekend. I am currently studying for my college physics midterm and got hung up on one of the problems that was on an old midterm.

The problem is as follows: Point charges Q1 = -5uC(micro C) and Q2 = +5.5 uC(micro C) are placed 0.4m apart as shown. The picture shows a 0.4 meter line pointing East, with the positive charge(Q2) on the right and the negative charge(Q1) on the left. Right above Q1, a 0.3 meter line goes North and ends at a point called P.

The drawing is pretty much a right triangle without the hypotenuse line drawn in.

The question wants me to calculate both the x and y components of the total electric field vector at point P. I have one vector line going South and the other going Northwest because of the positive charge. The problem does not give any angles so I'm not sure if they want me to find them or not?

Should I use E = kq/r^2? or Coulomb's Law? I greatly appreciate people for looking at this. I know most of the other problems on this midterm except this one!
 
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  • #2
Actually just figured it out myself. I learned to get the magnitudes of the electric fields first, then you bet I needed the angles for the x and y components. The tricky part was drawing and visualizing the vector angles the correct way, meaning which way i wanted sin or cos to be directed from.

Just in case anyone looked at it, the answers are below.

For magnitude of E1(the 0.3 m) I got 499444.4444. For magnitude of E2(I had to find the hyp of .5 for the r), I got 179800.

I found the bottom right angle of 36.86 deg, and the top north west angle of 53.13 deg. sin53.13 deg * 499444.4444 = -3.99X10^5(direction is down on the y axis)
cos36.86 deg * 179800 = -1.4X10^5(direction is left on the x axis)

Phew that feels good!
 
  • #3
Besides, wrong forum mate.
 

What is an electric field vector?

An electric field vector is a mathematical representation of the direction and strength of the electric field at a specific point in space. It is a vector quantity, meaning it has both magnitude and direction.

How do I calculate electric field vectors at a point called P?

To calculate the electric field vector at point P, you will need to know the charge of the source creating the electric field, the distance between the source and point P, and the direction of the electric field. You can use Coulomb's law or Gauss's law to determine the magnitude of the electric field, and then use vector addition to determine the direction of the electric field at point P.

What units are used to measure electric field vectors?

Electric field vectors are typically measured in units of Newtons per Coulomb (N/C) or Volts per meter (V/m).

What factors affect the strength of an electric field at a point?

The strength of an electric field at a point is affected by the magnitude of the charge creating the field, the distance from the source of the field, and the medium through which the field is passing. In a vacuum, the strength of an electric field decreases with the square of the distance from the source, while in a medium such as air, the strength of the field may be affected by the dielectric constant of the medium.

Why is it important to calculate electric field vectors at a specific point?

Knowing the electric field at a specific point is important because it allows us to predict the behavior of charged particles in that region. This information is essential in many applications, such as designing electronic circuits and understanding the behavior of lightning strikes.

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