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Electric field strength at the center of an equilateral triangle

  • #1

Homework Statement



3 rods each of length 1 meter form an equilateral triangle. Two rods have a uniform charge distribution of [tex]8\times 10^{-6}[/tex] C and the third of [tex]-8\times 10^{-6}[/tex] C. What is the electric field strength at the center of an equilateral triangle

Homework Equations



[tex]\vec{E} = \frac{8.99\times 10^{9} \times q}{r^{2}}[/tex]

The Attempt at a Solution



I worked out the magnitude of the electric field strength at the center due to only one of the rods.

I got the following integral:

[tex]\vec{E} = 2\times 8.99\times 10^{9}\int_{0}^{l}(\frac{y\lambda dx}{(x^{2}+y^{2})^{\frac{3}{2}}})\widehat{\vec{y}} = \frac{2\times 8.99\times 10^{9}\lambda l}{y\sqrt{y^{2}+l^{2}}}\widehat{\vec{y}}[/tex]

After that I substituted the values given in the question and got:

[tex]\vec{E} = \frac{2\times 8.99\times 10^{9}\times 8\times 10^{-6}\times 0.5}{(\frac{0.5}{\sqrt{3}})\sqrt{(\frac{0.5}{\sqrt{3}})^{2}+(0.5)^{2}}} = 431520[/tex]

Since the horizontal components of the positively charged rods are equal but opposite, they cancel out so we have the following:

[tex]\vec{E} = (2\times 431520 \times sin(30))+(431520)) = 863040 N/C[/tex]

But this answer is wrong.

I have been trying to get this question done since yesterday and never got the right answer. Would very much appreciate your help.
 

Answers and Replies

  • #2
gneill
Mentor
20,795
2,773

Homework Statement



3 rods each of length 1 meter form an equilateral triangle. Two rods have a uniform charge distribution of [tex]8\times 10^{-6}[/tex] C and the third of [tex]-8\times 10^{-6}[/tex] C. What is the electric field strength at the center of an equilateral triangle

Homework Equations



[tex]\vec{E} = \frac{8.99\times 10^{9} \times q}{r^{2}}[/tex]

The Attempt at a Solution



I worked out the magnitude of the electric field strength at the center due to only one of the rods.

I got the following integral:

[tex]\vec{E} = 2\times 8.99\times 10^{9}\int_{0}^{l}(\frac{y\lambda dx}{(x^{2}+y^{2})^{\frac{3}{2}}})\widehat{\vec{y}} = \frac{2\times 8.99\times 10^{9}\lambda l}{y\sqrt{y^{2}+l^{2}}}\widehat{\vec{y}}[/tex]

After that I substituted the values given in the question and got:

[tex]\vec{E} = \frac{2\times 8.99\times 10^{9}\times 8\times 10^{-6}\times 0.5}{(\frac{0.5}{\sqrt{3}})\sqrt{(\frac{0.5}{\sqrt{3}})^{2}+(0.5)^{2}}} = 431520[/tex]

Since the horizontal components of the positively charged rods are equal but opposite, they cancel out so we have the following:

[tex]\vec{E} = (2\times 431520 \times sin(30))+(431520)) = 863040 N/C[/tex]

But this answer is wrong.

I have been trying to get this question done since yesterday and never got the right answer. Would very much appreciate your help.
Hi Alexander2357, Welcome to Physics Forums.

Your solution method looks okay to me. If you are entering the results into an automated marking system perhaps you are getting "bitten" by the formatting? Make sure that your significant figures agree with those of your given values and constants.
 
  • #3
Simon Bridge
Science Advisor
Homework Helper
17,856
1,654
Note:
Niggle: You are missing units of pretty much every result except the last one.

The strategy looks good.
I think you may have made some arithmetic errors.
You should do all the algebra before plugging the numbers in.
I think the equation ends up taking a particularly simple form.

If you check - the answer should be about the order of ##k\lambda## shouldn't it? (since y and l are order 10^0) Roughly 10^4? Yours is order 10^5.

The two positive rods provide field vectors with 120deg between them - you can get the resultant by the cosine rule. This resultant ends up in the same direction as the contribution from the negative rod.

Like gniell says, make sure the answer you submit fits the expected format.
 

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