What is the electric field intensity at point A due to two point charges?

In summary: The distance from q2 to point A can be found using trigonometry and the given values. The angle that the field from q2 makes with the horizontal can also be found using trigonometry. Once you have the magnitude and direction of both E1 and E2, you can use the component method of vector addition to find the resultant electric field E at point A.
  • #1
Inertialforce
68
2

Homework Statement


Find Enet (magnitude and direction) at point A and include a vector diagram.

q1 = 2.0x10^-6 C
q2 = 6.0x10^-6 C
r1 = 2.5x10^-3 m
r2 = 1.7x10^-3 m

Homework Equations


E = Klql/r^2 or E = Kq/r^2 (the first equation just has absolute values around the q for reminder but they both are the same thing)


The Attempt at a Solution


First I looked at the diagram and I know that the electric field created by a charge will always put away from a positive charge and towards a negative so I drew a vector diagram for that at point "A" using tip to tail method, then I joined E1(electric field created by q1) and E2(electric field created by q2) to have a resultant of E.

I then used the equation E = kq/r^2 to find out what E1 and E2 are and I found out that they equal 2.8x10^9 and 1.9x10^10 respectively. My question, now I have to use the component method of vector addition to find out what E is but I need an angle to do that. Is the angle in this case 90 degrees?
 

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  • #2
Inertialforce said:
I then used the equation E = kq/r^2 to find out what E1 and E2 are and I found out that they equal 2.8x10^9 and 1.9x10^10 respectively.
How did you find E2? What value of r (or r^2) did you use?
My question, now I have to use the component method of vector addition to find out what E is but I need an angle to do that. Is the angle in this case 90 degrees?
No. Figure out the angle that E2 makes with the horizontal by examining that triangle and using a little trig.
 
  • #3
Doc Al said:
How did you find E2? What value of r (or r^2) did you use?

I figured out E2 by using the values q2 = 6.0x10^-6C and r(or r^2) = 1.7x10^-3m. Because these values were given at the start of the question, we didn't need to find them.

Then I took those numbers and I directly plugged it into the equation E= kq/r^2.
 
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  • #4
Inertialforce said:
I figured out E2 by using the values q2 = 6.0x10^-6C and r(or r^2) = 1.7x10^-3m. Because these values were given at the start of the question, we didn't need to find them.

Then I took those numbers and I directly plugged it into the equation E= kq/r^2.
To figure out E2 (the field from q2 at point A) you need the distance from q2 to point A, not the distance between the charges (which is given as 1.7x10^-3m).
 

1. What is Electric Field Intensity?

Electric Field Intensity, also known as Electric Field Strength, is a measure of the strength of an electric field at a particular point in space. It is a vector quantity, meaning it has both magnitude and direction, and is measured in units of volts per meter (V/m).

2. How is Electric Field Intensity calculated?

Electric Field Intensity is calculated by dividing the force experienced by a test charge placed in an electric field by the magnitude of the test charge. Mathematically, it can be expressed as E = F/q, where E is the electric field intensity, F is the force experienced by the test charge, and q is the magnitude of the test charge.

3. What factors affect Electric Field Intensity?

The factors that affect Electric Field Intensity include the magnitude of the source charge, the distance between the source charge and the point in space where the electric field is being measured, and the medium in which the electric field exists. Other factors such as the presence of other charges and the shape of the source charge can also have an influence.

4. How is Electric Field Intensity different from Electric Potential?

Electric Field Intensity and Electric Potential are related but different concepts. Electric Field Intensity measures the strength of the electric field itself, while Electric Potential measures the potential difference between two points in an electric field. Electric Potential is a scalar quantity, meaning it has only magnitude and no direction, and is measured in units of volts (V).

5. What are some real-life applications of Electric Field Intensity?

Electric Field Intensity has many practical applications, including in the design of electronic devices, in the study of electrostatics and electromagnetics, and in the development of technologies such as electric motors, generators, and capacitors. It is also used in medical procedures, such as electrocardiograms, and in the study of atmospheric electricity and lightning.

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