Magnitude and direction of electric force (file attached)

In summary, the question is about determining the radius for the charge on q3 by q1, which is found to be r^2 = 2a^2 using the Pythagorean theorem. The notes also discuss the introduction of acceleration in electric fields and solving for the value of time in a particular equation. The issue with the cosine function is clarified by noting the change in angle from \theta to 2\theta.
  • #1
trah22
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Homework Statement


I was kind of confused on how the radius for the charge on q3 by q1 is 2a^2, i attached the file for my notes regarding this. The question is at the bottom of the pdf scan.


Homework Equations





The Attempt at a Solution

 

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  • #2
The needed distance is the hypotenuse of the right triangle whose sides are both "a". Hint: Pythagorean theorem.
 
  • #3
c^2=a^2+b^2

so

c^2=a^2+a^2
c^2=2a^2

what about the ^2 for the c, is it just left there? because it seems like it since r=2a^2
 
  • #4
Careful: it's r^2 = 2a^2, not r = 2a^2.
 
  • #5
On this other page of notes which has to do with the introduction of acceleration in electric fields, there are three parts of the problem that I am not quite following, i figured it would be easier if i just wrote it down by hand on the actual note sheet than typing it all out.:smile:

For the cosign question, i don't undestand what happened to it because the sin is still there in the next step but not the cos.
 

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  • #6
If what I am asking isn't really quite clear, i can write it out again more specifically and scan it.
 
  • #7
(Be sure to include the full problem statement next time.)

(1) Why negative? The field points down (thus is negative), the charge is positive: so the force and acceleration are both downward, thus negative.

(2) They are solving for the value of time that makes y = 0. Set that factor equal to zero and solve for t.

(3) Regarding what happened to the cosine: note that [itex]\theta[/itex] changed to [itex]2\theta[/itex]. (Review your trig identities for [itex]\sin\theta \cos\theta[/itex].)
 

FAQ: Magnitude and direction of electric force (file attached)

What is electric force?

Electric force is the force exerted between two charged particles or objects. It is a fundamental force of nature and is responsible for the interaction between particles that have an electric charge.

How is the magnitude of electric force calculated?

The magnitude of electric force can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula for calculating electric force is F = k(q1q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

What is the unit of electric force?

The unit of electric force is Newton (N), which is the same as the unit for any type of force. In the SI (International System of Units) system, 1 N is equal to 1 kg*m/s^2.

Does electric force have direction?

Yes, electric force has both magnitude and direction. The direction of electric force is determined by the sign of the charges involved. If the charges are of the same sign, the force is repulsive and if they are of opposite signs, the force is attractive.

How does distance affect the electric force between two charged particles?

The electric force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases. For example, if the distance between two particles is doubled, the force between them will decrease by a factor of 4.

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