Why Am I Getting the Wrong Electric Field Intensity Calculation?

In summary, the conversation discusses a problem with finding the field intensity at a specific point on a rectangle with sides 6cm and 8cm. The point charges and their values are given, and the method of finding the field intensity is discussed using the equation E = kq/r^2. The steps involved in finding the field intensity are broken down and the desired answer is given as 6.93 * 10^3 at an angle of 78.8 under the horizontal. The conversation ends with a request for help in finding where the error is in the calculations.
  • #1
bluntz48
1
0
i have been stuck on this problem for days and can't figure out where I'm going wrong. i have a rectangle with sides 6cm and 8cm. 3 points of the triangle have a point charge on them with the following values:

q1 = -2 * 10^(-8) C
q2 = 8 * 10^(-8) C
q3 = -4 * 10^(-8) C

The point charges are arranged such that q1 is on top right, q2 on top left, q3 on bottom left. Find the field intensity at the 4th point (bottom right).

So what I did is use E = kq/r^2 (k = 9*10^9) to find the field intensity on the 4th point. For q2 I broke it into x- and y- components with E2x = E2 cos 36.87 and E2y = E2 sin 36.87. For q2 I found the angle of 36.87 by doing the inverse tan of 0.06/0.08 = 0.75 and i found the distance between the two points by using the pythagorean theorum (a^2 + b^2 = r^2).

Once I have both E1 and E3 as well as E2 broken into components, I add up the ones on the x-axis (E2x, E3) and the ones on the y-axis (E2y, E1). Then again I use the pythagoren theorum to add up the resultant (Ex^2 + Ey^2 = R^2)

I keep getting the wrong answer and I don't know why. Could you please point out where I'm going wrong or list the steps I should be taking so I could compare. The answer on the sheet is 6.93 * 10^3 at an angle of 78.8 under the horizontal

thanks everyone!
 
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  • #2
bluntz48 said:
i have been stuck on this problem for days and can't figure out where I'm going wrong. i have a rectangle with sides 6cm and 8cm. 3 points of the triangle have a point charge on them with the following values:

q1 = -2 * 10^(-8) C
q2 = 8 * 10^(-8) C
q3 = -4 * 10^(-8) C

The point charges are arranged such that q1 is on top right, q2 on top left, q3 on bottom left. Find the field intensity at the 4th point (bottom right).

So what I did is use E = kq/r^2 (k = 9*10^9) to find the field intensity on the 4th point. For q2 I broke it into x- and y- components with E2x = E2 cos 36.87 and E2y = E2 sin 36.87. For q2 I found the angle of 36.87 by doing the inverse tan of 0.06/0.08 = 0.75 and i found the distance between the two points by using the pythagorean theorum (a^2 + b^2 = r^2).

Once I have both E1 and E3 as well as E2 broken into components, I add up the ones on the x-axis (E2x, E3) and the ones on the y-axis (E2y, E1). Then again I use the pythagoren theorum to add up the resultant (Ex^2 + Ey^2 = R^2)

I keep getting the wrong answer and I don't know why. Could you please point out where I'm going wrong or list the steps I should be taking so I could compare. The answer on the sheet is 6.93 * 10^3 at an angle of 78.8 under the horizontal

thanks everyone!

Sounds like your approach is right. The rectangle is 6 high by 8 wide?

Can you list the x and y components of E that you get for each of the 3 sources, so that we can check your math? E1 just points up, and E3 just points left, and E2 points down and to the right, correct?
 
  • #3


I would first like to commend you for your efforts in trying to solve this electric field problem. It can be frustrating when we get stuck on a problem for days, but it's important to remember that persistence and critical thinking are key in solving scientific problems.

From your explanation, it seems like you have a good understanding of the formula for electric field intensity and how to break down components. However, I would suggest double checking your calculations and making sure you are using the correct values for distances and angles. It's always a good idea to go back and check your work to catch any mistakes.

Another approach you could try is using vector addition to find the resultant electric field at the 4th point. This involves adding the individual electric field vectors, taking into account their direction and magnitude. This method may help you avoid any errors in breaking down components.

Additionally, it's important to note that the angle of 36.87 degrees for q2 may not be correct. It's always a good idea to draw a diagram and label all the given values to avoid any confusion.

I hope this helps guide you in the right direction. Remember to always double check your calculations and approach the problem from different angles if you're having trouble. Good luck!
 

Related to Why Am I Getting the Wrong Electric Field Intensity Calculation?

1. What is an electric field?

An electric field is a physical phenomenon that describes the influence of electrically charged objects on each other. It is a vector quantity that represents the direction and strength of the force exerted on a charged particle at any given point in space.

2. How is an electric field calculated?

The electric field at a point in space is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. This is known as Coulomb's law, which states that the electric 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 the units of an electric field?

The SI unit of electric field is newtons per coulomb (N/C). However, it can also be expressed in volts per meter (V/m) or even electronvolts per meter (eV/m) in some cases.

4. How does an electric field affect charged particles?

An electric field exerts a force on any charged particles within its influence. Positively charged particles will be pushed in the direction of the electric field, while negatively charged particles will be pushed in the opposite direction.

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

Electric fields have many practical applications, such as in electrical circuits, generators, and motors. They are also used in technologies such as particle accelerators, cathode ray tubes, and capacitors. In nature, electric fields play a crucial role in the functioning of living organisms, such as in the nervous system and muscle contractions.

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