Strength and Direction of Electric Field

In summary, the conversation is about a problem involving the calculation of the strength and direction of the electric field at a specific position. The person asking the question provides their attempted solution, asking for help in finding any potential mistakes. The expert summarizer confirms that the work looks correct, but suggests to check for round-off error. The person then discovers that a comma in their vector notation was causing the issue, and thanks the expert for their help.
  • #1
TheCarl
21
0

Homework Statement



What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form.
27.P28.jpg


Homework Equations



E=kQ/r^2

The Attempt at a Solution



E(+5) = (8.99E9)(5E-9)/(0.02^2) = (1.1E5i, 0j)
E(-5) = (8.99E9)(5E-9)/(0.02^2) = (0i, 2.8E4j)
r(+10) = SQR(0.02^2+0.04^2) = 0.0447
E(+10) = (8.99E9)(10E-9)/(0.0447^2) = 4.5E4
arctan(2/4)=26.57DEG
= (4.5E4sin(26.57)i, 4.5cos(26.57)j)
= (2.0E4i, -4.0E4j)​

E(+5) + E(-5) + E(10) = (1.3E5i, -1.2E4j)

Mastering Physics is telling me that this answer is wrong and I want to know where I screwed up.
 
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  • #2
Your work looks good to me. I don't see any mistake.
 
  • #3
TheCarl said:

Homework Statement



What are the strength and direction of the electric field at the position indicated by the dot in the figure? Answer in component form.
[ IMG]http://session.masteringphysics.com/problemAsset/1384377/2/27.P28.jpg[/PLAIN]

Homework Equations



E=kQ/r^2

The Attempt at a Solution



E(+5) = (8.99E9)(5E-9)/(0.02^2) = (1.1E5i, 0j)
E(-5) = (8.99E9)(5E-9)/(0.02^2) = (0i, 2.8E4j)
r(+10) = SQR(0.02^2+0.04^2) = 0.0447
E(+10) = (8.99E9)(10E-9)/(0.0447^2) = 4.5E4
arctan(2/4)=26.57DEG
= (4.5E4sin(26.57)i, 4.5cos(26.57)j)
= (2.0E4i, -4.0E4j)​

E(+5) + E(-5) + E(10) = (1.3E5i, -1.2E4j)

Mastering Physics is telling me that this answer is wrong and I want to know where I screwed up.
The problem may well be too much round-off error.

Keep at least two extra digits for all intermediate results.

Round-off your final result to the correct sig. dig. if whoever/whatever is grading your work is anal about such things.
 
  • #4
Turns out it was the comma in my vector notation. Once I removed the comma, the computer accepted it. Looks like I'm a bit rusty on that stuff. Thank you both for your help.
 
  • #5


I would like to provide a few suggestions for your solution:

1. Make sure you are using the correct formula for electric field strength. The equation you provided, E = kQ/r^2, is the correct one for point charges, but it looks like the figure in the problem is showing multiple point charges, so you may need to use the superposition principle to calculate the total electric field at the position indicated by the dot.

2. Check your units. The electric field strength is typically measured in units of N/C (newtons per coulomb), so make sure your final answer has the correct units.

3. When calculating the electric field at the position indicated by the dot, you need to take into account the direction of each individual electric field. Simply adding the components of each electric field may not give you the correct answer. Instead, you may need to use vector addition to find the total electric field at that point.

4. Double check your trigonometry. The angle you calculated using arctan(2/4) is not correct. Remember, tangent is opposite over adjacent, so in this case it would be 4/2 = 2, not 2/4. Also, be careful with your signs when using trig functions - in this case, you should be using cos(26.57) = 0.894 and sin(26.57) = 0.448, not the negative values you have in your solution.

I hope these suggestions help you find the mistake in your solution. As a scientist, it's important to always double check your work and make sure you are using the correct equations and units. Good luck!
 

Related to Strength and Direction of Electric Field

1. What is the strength of an electric field?

The strength of an electric field is measured by the force experienced by a unit charge placed in the field. It is expressed in units of Newtons per Coulomb (N/C).

2. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction of the force it exerts on a positive test charge. The direction of the field is always pointing away from positive charges and towards negative charges.

3. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the source charge, the distance from the source charge, and the medium through which the field is passing. The strength of the field decreases with distance and also depends on the relative permittivity of the medium.

4. How is the strength of an electric field represented graphically?

The strength of an electric field is represented graphically by electric field lines. These lines show the direction of the field and are drawn closer together in areas of higher field strength. The number of lines per unit area is also a representation of the field strength.

5. Can the direction and strength of an electric field be changed?

Yes, the direction and strength of an electric field can be changed by altering the source charge or the distance from the source charge. The use of conductors or insulators also affects the strength and direction of an electric field.

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