Electrostatics - Point charge question

In summary, you used the formula E = k \frac{q}{r^2} to find the electric field strength of the two positive charges along the square edge as 7.39 \times 10^6 N/C. For the positive charge along the diagonal, you used the diagonal of the square as the value for "r" and resolved the resulting electric field strength into x and y components. The presence of the negative charge does not affect the magnitude of the electric field due to the other 3 positive charges.
  • #1
joannananaa
2
0
1. Three equal positive point charges of magnitude Q = 5.00μ C are located at three corners of a square of edge length d = 7.8 cm. A negative charge -15.00μ C is placed on the fourth corner. At the position of the negative charge, what is the magnitude of the electric field due to the three positive charges?



2. Homework Equations :
F=KQ/R^2


3. The Attempt at a Solution :
I have used the equation K=KQ/R^2 to find that the positive points of charge are 7.4E6 N. I wrote these as <7.4E6, 0> and <0, -7.4E6>. I used this equation on the negative charge as well, and got 11.2E6, which I resolved into X and Y components, which were both 7.9E6 and wrote these as <7.9E6, -7.9E6>.

All of this is just following notes from class, and I am unsure of how to actually get the answer.

Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2
welcome to pf!

hi joannananaa! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
joannananaa said:
I have used the equation K=KQ/R^2 to find that the positive points of charge are 7.4E6 N. I wrote these as <7.4E6, 0> and <0, -7.4E6>. I used this equation on the negative charge as well, and got 11.2E6, which I resolved into X and Y components, which were both 7.9E6 and wrote these as <7.9E6, -7.9E6>.

i'm confused :redface:

there are 3 positive charges, 2 at distance d, and 1 at distance d√2 …

how did you get your figures? :confused:
 
  • #3
Thank you :)

Um...I just plugged them into the formula and wrote one as negative and one as positive. I'm pretty lost.
 
  • #4
The presence of the negative charge is irrelevant to the magnitude of electric field due to the other 3 positive charges.
Your E field strength of [tex]7.39 \times 10^6 [/tex] N/C for the two charges along the square edge was right.
However, for the one along the diagonal, your "r" in [tex]E = k \frac{q}{r^2}[/tex] needs to be the diagonal of the square.
Then you just resolve this into x and y components, and add all 3 electric field strengths as vectors to get the E field strength in the 4th corner.
 
  • #5


Based on the information provided, the magnitude of the electric field at the position of the negative charge can be calculated using the superposition principle. This principle states that the total electric field at a point due to multiple point charges is equal to the vector sum of the individual electric fields at that point.

Using this principle, we can calculate the electric field at the position of the negative charge by considering the electric fields due to each of the three positive charges separately. The electric field due to a single point charge is given by the equation E = KQ/R^2, where K is the Coulomb's constant, Q is the magnitude of the charge, and R is the distance from the charge to the point where the electric field is being calculated.

In this case, the distance R from each of the three positive charges to the position of the negative charge is equal to the diagonal of the square, which can be calculated using the Pythagorean theorem as d/√2 = 7.8 cm/√2 = 5.5 cm. Therefore, the electric field due to each of the three positive charges is given by E = (9x10^9 Nm^2/C^2)(5.00 μC)/(5.5 cm)^2 = 5.01x10^5 N/C.

Since the electric fields due to the three positive charges are all acting in the same direction (towards the negative charge), we can add them together using vector addition. This gives us a total electric field of E = 3(5.01x10^5 N/C) = 1.50x10^6 N/C at the position of the negative charge.

Therefore, the magnitude of the electric field at the position of the negative charge is 1.50x10^6 N/C.
 

1. What is a point charge?

A point charge is an electric charge that is concentrated at a single point in space and has no physical size. It is often represented by the symbol "q" and can be either positive or negative.

2. How is the strength of a point charge determined?

The strength of a point charge is determined by its magnitude, which is measured in coulombs (C). This magnitude is a measure of the amount of electric charge the point charge possesses.

3. What is the relationship between distance and electric force for a point charge?

The electric force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. This relationship is described by Coulomb's Law: F = k(q1q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the magnitudes of the two point charges, and r is the distance between them.

4. Can the direction of the electric force on a point charge be changed?

Yes, the direction of the electric force on a point charge can be changed by the presence of another point charge. The electric force is always attractive for opposite charges and repulsive for like charges.

5. What is the difference between electric potential and electric potential energy for a point charge?

Electric potential is a measure of the potential energy per unit charge at a given point in an electric field. It is measured in volts (V). On the other hand, electric potential energy is the energy that a point charge possesses due to its position in an electric field. It is measured in joules (J).

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
635
  • Introductory Physics Homework Help
Replies
21
Views
604
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
752
Replies
17
Views
878
  • Introductory Physics Homework Help
Replies
17
Views
323
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
730
  • Introductory Physics Homework Help
Replies
9
Views
976
  • Introductory Physics Homework Help
Replies
11
Views
622
Back
Top