Electrostatics problem (ruddy angles)

In summary: The distance from the 1.5 uC charge to P is 7.31 cm, and the distance from the 9.2 uC charge to P is 4.69 cm. The fields from each charge are not just the ones you've calculated from 7.31 cm and 4.69 cm. The total field at P is the vector sum of the two fields.In summary, the magnitude of the field strength at point P, given a distance of 5.6 cm between the centers of 1.5E-6 C and 9.2E-6 C charges, can be calculated using the equations E = kq/r^2 and E = F/Q. However
  • #1
Destrio
212
0

Homework Statement


Calculate the magnitude of the field strength at point P as shown. The distance between the centers of the 1.5E-6 C and the 9.2E-6 C charges is 5.6 cm..


Homework Equations


E = kq/r^2
E = F/Q
E = V/d


The Attempt at a Solution


Used tan and sin of 50 to find all distances

E = kq/r^2
E = (9E9)(1.5E-6C) / (7.31E-2 m)^2
E = 2.526E6 N/C

E = kq/r^2
E = (9E9)(9.2E-6C) / (4.69E-2 m)^2
E = 2.526E6 N/C

then used those numbers to find c^2 using cosine law.

I'm not getting a correct answer (obvoiusly).

Thanks,
 

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  • #2
picture is now working
 
  • #3
What's a c^2 cosine law? You have the two vector components of the total E field. Do you know how to add the components? Is the answer expressed in polar or rectangular form?
 
  • #4
cosine law is:
c^2 = a^2 + b^2 - 2ab cos c
I don't know how to add the components or what polar/rectangular form is
 
  • #5
Destrio said:

Homework Statement


Calculate the magnitude of the field strength at point P as shown. The distance between the centers of the 1.5E-6 C and the 9.2E-6 C charges is 5.6 cm..


Homework Equations


E = kq/r^2
E = F/Q
E = V/d


The Attempt at a Solution


Used tan and sin of 50 to find all distances

E = kq/r^2
E = (9E9)(1.5E-6C) / (7.31E-2 m)^2
E = 2.526E6 N/C

E = kq/r^2
E = (9E9)(9.2E-6C) / (4.69E-2 m)^2
E = 2.526E6 N/C

then used those numbers to find c^2 using cosine law.

I'm not getting a correct answer (obvoiusly).

Thanks,

There is no way the fields from the two charges are the same. Your distances look OK.
 

Related to Electrostatics problem (ruddy angles)

1. What is electrostatics problem (ruddy angles)?

Electrostatics problem (ruddy angles) is a physics problem that deals with the forces and interactions between electrically charged objects at different angles.

2. How is electrostatics problem (ruddy angles) solved?

The electrostatics problem (ruddy angles) is solved by using the principles of Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What are ruddy angles in electrostatics problem?

Ruddy angles refer to the angles at which two charged objects are positioned in relation to each other in an electrostatics problem. These angles are important in determining the magnitude and direction of the electrostatic force between the objects.

4. How does the distance between charged objects affect electrostatics problem (ruddy angles)?

The distance between charged objects has a significant impact on the electrostatics problem (ruddy angles). As the distance between the objects increases, the electrostatic force between them decreases, and therefore the ruddy angles will also change.

5. What are the real-world applications of electrostatics problem (ruddy angles)?

Electrostatics problem (ruddy angles) has various real-world applications, including the design and construction of electrical circuits, the behavior of particles in an electric field, and the study of the forces between charged particles in space.

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