charges on the vertex of a triangle


by Rasine
Tags: charges, triangle, vertex
Rasine
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#1
May16-07, 07:10 PM
P: 214
Positive point-charges of +16.0 mC are fixed at two of the vertices of an equilateral triangle with sides of 1.30 m, located in vacuum. Determine the magnitude of the E-field at the third vertex.

so...all the particals are in equalibrum so that means F=ma=0

so if i call the third vertex c...Ea+Eb+Ec=0 so that means that Ea+Eb=-Ec

now i can write kqa/r^2+kqb/r^2=Ec and i just have to solve for Ec right?
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mjsd
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May16-07, 07:34 PM
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note that E-fields are vectorial quantities. Ie. they have a direction. in your case, two point charges at two different location each exerting a Force radially, will induce two different field vectors at the the third spot. These two vectors are not parallel and so need to take the angle in between into account to get the right answer.

btw, the question doesn't seem to mention anything about the system being in equilibrium, just asking you work out the resulant field at third spot
Rasine
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#3
May16-07, 07:44 PM
P: 214
ummm...so do i have to find in they x and y compnents and all that sutff?

jackiefrost
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#4
May16-07, 10:21 PM
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charges on the vertex of a triangle


Quote Quote by Rasine View Post
Positive point-charges of +16.0 mC are fixed at two of the vertices of an equilateral triangle with sides of 1.30 m, located in vacuum. Determine the magnitude of the E-field at the third vertex.

so...all the particals are in equalibrum so that means F=ma=0

so if i call the third vertex c...Ea+Eb+Ec=0 so that means that Ea+Eb=-Ec

now i can write kqa/r^2+kqb/r^2=Ec and i just have to solve for Ec right?
Try defining your coordinate system so you can locate the two known charges at -0.65m and +0.65m on the x axis and then place third vertex appropriately on the y axis. This has a nice benefit. Since your two charges are equal and of the same polarity, the x components of the E fields that they produce at the third vertex will exactly cancel. So, you just need to find the y components and add them. The kQa/r^2 and kQb/r^2 are the magnitudes of the two E vectors but don't forget - they are pointing in different directions so you need to compute the y components according to their angles' with the y axis.

Don't panic - it's simple trig - sketch the two vectors at the 3rd vertex point keeping in mind that the E field for each charge is defined in terms of the force per unit positive test charge; i.e. the force which would be produced on a positive unit charge placed at the point under consideration. For your case, each charge would repel a unit positive test charge - notice the direction each would be trying to repel it. They're NOT the same direction. So, as you can see (if you really sketched the situation), you must find the vector sum of the two E vectors which, in this case, ends up being just the sum of their y components.

BTW - this is a static problem - there's no motion involved. It's like an ideal, instantaneous snapshot just before any motion could take place.

jf


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