Calculating Electric Field at a Given Position Due to a π- Particle

In summary, the conversation discusses the process of finding the electric field at a given location due to a π- particle. The steps involve subtracting the source location from the observation location, calculating the distance between the two points, and using the electric field equation to find the magnitude and direction of the electric field. After some initial errors, the final answer is determined to be <2.997e6, -2.397e6, -3.296e6> N/C.
  • #1
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Homework Statement



A π- ("pi-minus") particle, which has charge -e, is at location ‹ 5.00e-9, -5.00e-9, -6.00e-9 › m. What is the electric field at location < -5.00e-9, 3.00e-9, 5.00e-9 > m, due to the π- particle?
5e-9 is equivalent to 5*10^-9

Homework Equations


E=(k*q/r^2)r-hat


The Attempt at a Solution


So to find the Electric field created at the given position first I subtract the source from the observation, I am assuming the pi- particle is the source. When I do this is what I get
<-10e-9,8e-9,11e-9>

So next I compute the r^2 or the distance from source to observation location. To do this I square each entry in the vector above and add them up to get 8.5e-17

Now I compute the value to the left of the r-hat vector: 9e9*(-1.602e-19/8.5e-17), this gives me -16962352.94 N/C

Next I calculate the r-hat vector by dividing r by its magnitude which is sqrt(8.5e-17).
I get <-1.0846, .8677, 1.1931>

[-16962352.94 N/C *<-1.0846, .8677, 1.1931>]
Finally I multiply the two numbers together and get a vector with huge values, which doesn't seem right, can some one tell me where I am going wrong or maybe tell me the right way to approach this problem, because I seem to be stuck.
Thank You for your help.
 
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  • #2
Your value for |r|2 looks dubious. Better check it.
 
  • #3
Yes the |r|^2 was dubious

After reworking that I got |r|=1.6882e-8

Plugging this back in I get
E=[9e9*(1.602e-19)/(1.6882e-8)^2]*< -0.5923, 0.4739, 0.6516>

=<2996700,-2397300,-3296300>

Is this a more reasonable answer?
 
  • #4
The results look good.

You might want to use scientific notation and appropriate significant figures for your final answer.
 
  • #5




Your approach is correct, but there are a few errors in your calculations. First, when you subtract the source from the observation, your result should be < -1.0e-8, 8.0e-9, 11.0e-9 > m, not < -10e-9, 8e-9, 11e-9 > m. This is because you are subtracting the coordinates of the source from the coordinates of the observation, not the other way around.

Secondly, when you calculate the magnitude of the displacement vector (r), you should get 1.3e-8 m, not 8.5e-17 m. This is because you need to take the square root of the sum of the squares of the components of the displacement vector, not the displacement vector itself.

Finally, when you calculate the electric field using the formula E = (k*q/r^2)r-hat, you should get a vector with values of -1.3e8 N/C, 1.0e8 N/C, and 1.4e8 N/C for the x, y, and z components respectively.

I hope this helps clarify the problem for you. Keep in mind to always double check your calculations and make sure you are using the correct units. Good luck with your studies!
 

FAQ: Calculating Electric Field at a Given Position Due to a π- Particle

What is an electric field?

An electric field is a region around a charged particle or object where a force is exerted on other charged particles or objects. It is represented by electric field lines, which show the direction and strength of the field.

How is the electric field calculated?

The electric field is calculated by dividing the force exerted on a test charge by the magnitude of the charge. This is expressed in the formula E = F/q, where E is the electric field, F is the force, and q is the charge of the test particle.

What is the unit of measurement for electric field?

The unit of electric field is Newtons per Coulomb (N/C) in the SI system. In other systems, it may also be expressed in volts per meter (V/m).

How does distance affect the strength of an electric field?

The strength of an electric field is inversely proportional to the square of the distance from the charged object. This means that the further away from the object, the weaker the electric field will be.

What are some real-life applications of electric field calculations?

Electric field calculations are used in a variety of real-life applications, such as in the design of electrical circuits, the operation of electronic devices, and the development of medical equipment. They are also important in understanding the behavior of lightning, the Earth's magnetic field, and the interaction between charged particles in outer space.

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