How Do You Calculate Electric Field Intensity?

In summary, the question asks for an equation that describes the effect of an imagined point positive charge on an electric field. The student provides an equation that describes the effect of an imagined point positive charge on an electric field. The student explains that the equation should be written in terms of the electric fields' vectors and that the computer is incorrect in marking the student's answer as correct.
  • #1
Ush
97
0

Homework Statement



The question, along with my attempt is in the image attached. P

Homework Equations



E = KQ/r2
E = electric field
Q = source charge
r = distance

The Attempt at a Solution



see document attached

----
thank you!
 

Attachments

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  • #2
Well I haven't run through the numbers, but theoretically it seems possible.
The electric field at whatever point is what would act on an imagined point positive charge, so with that in mind:

You have q1 pulling the charge left, and q3 pushing it also left. q2 has no horizontal affect.

Then you've done the nice Pythagoras stuff and found that q1 pulls the charge down, but q2 and q3 push it upwards. Since q2 + q3 have a greater magnitude than q1, they will win.

Pythagoras-ing horizontal and vertical components gives your perfectly depicted arrow, and a magnitude that agrees in sign with this hand waving argument.
Was there something you didn't understand?
 
Last edited:
  • #3
Onamor said:
Was there something you didn't understand?

to me, it makes sense. But to the computer marking my answer, it does not =[
 
  • #4
Hmm, do you know what the correct answer is?

I got a different answer to yours with just the superposition equation:
ddca45ddf0a04f30a7280a846389febf.png


It may just be a computational mistake - when you are calculating the components, you have them on the order nC (x10^-9 in SI units). But the Coulomb constant is
cimg336.gif

So they should not be of this order.
 

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  • #5
I checked over my calculations but i get the same number, I don't know what the correct answer is

I'm not sure how you did your superposition,
but that just looks like the equation for E, - the sum of the electric fields. Electric fields are vectors, you can't add them without breaking them into components- at least I don't know how to yet-
 
  • #6
pleaseeee someone help!?
 
  • #7
I didn't check your arithmetic, but your method looks correct.
 
  • #8
the computer keeps marking it wrong though =/ this seems like a simple question..
 
  • #9
Often the online systems are fussy about significant figures--some of them insist on 3 sig figs no matter what.
 
  • #10
gosh.. thanks o_o it worked!
 

1. What is an electric field?

An electric field is a physical field that is created by charged particles. It is a vector field, meaning it has both magnitude and direction, and it describes the influence that an electric charge has on other charged particles in its vicinity.

2. How is an electric field measured?

An electric field is typically measured using a device called an electric field meter. This meter measures the strength of the electric field at a specific point in space, and the result is usually given in units of volts per meter (V/m).

3. What is the difference between an electric field and an electric potential?

An electric field is a physical quantity that describes the force experienced by a charged particle in the presence of other charged particles. On the other hand, electric potential is a scalar quantity that describes the potential energy per unit charge at a specific point in space.

4. How can I calculate the electric field at a specific point?

The electric field at a specific point can be calculated using the equation E = kQ/r², where k is the Coulomb's constant, Q is the source charge, and r is the distance from the source charge to the point where the electric field is being measured. Alternatively, the electric field can also be calculated using the equation E = -∇V, where V is the electric potential.

5. What are some real-life applications of electric fields?

Electric fields have many practical applications, including powering electronic devices, generating electricity, and controlling the motion of charged particles in particle accelerators. They are also used in technologies such as electrostatic precipitators, which remove pollutants from industrial exhaust gases, and electrostatic air filters, which remove dust and other particles from the air.

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