An equilateral triangle's electric field at its center

[Roverse]

Homework Statement

Three 18-cm long rods form an equilateral triangle. Two of the rods are charged to +10 nC, and the third to - 10 nC.
What is the electric field strength at the center of the triangle?

Homework Equations

$$ \vec{E} = \frac{k*q}{r^2} $$

The Attempt at a Solution

1. Draw the thing.

2. Integrate.

$$\int_0^d\frac{\left(k\cdot\frac{Q}{d}\right)}{\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+\left(d-\left(\frac{d}{2}\right)^2\right)}}\cdot\left(\frac{\left(\left|\frac{d}{2}-x\right|\right)}{\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+d-\left(\frac{d}{2}\right)^2}}\right)dx$$

*d is in meters instead of centimeters

3. Add the vectors.

Because the charges are equal, and it is an equilateral triangle, I'm just going to simplify things.
The Y component is the bottom bar which is charge 10 with the magnitude of the integral above. The other two cancel.

The X component is 2*the integral of above from both the negative plate on the right and the positive plate on the left.

The magnitude is $$\sqrt{y^{2} + x^{2}}$$

For this problem the answer I computed is 0.0105401. I messed up somewhere and cannot find it...

For a sanity check, I know it must be less than the point charges at the center of the rods ~ sqrt(1.4^2+1.4cos(45)^2), but my answer seems too small.Thanks in advance.

SOLUTION:

If I change the length of the height to sqrt((d^2-(d/2)^2)/2 The integral does work: (Fixed)

$$\int_0^d\frac{\left(k\cdot\frac{Q}{d}\right)}{\left(\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+\left(\frac{\sqrt{d^2-\left(\frac{d}{2}\right)^2}}{2}\right)^2}\right)}\cdot\frac{\left(\left|\frac{d}{2}-x\right|\right)}{\left(\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+\left(\frac{\sqrt{d^2-\left(\frac{d}{2}\right)^2}}{2}\right)}\right)}dx$$

 

[kuruman]

What expression did you use for the field due to a charged rod at a point on its perpendicular bisector? That is of relevance here.

 

[Roverse]

What expression did you use for the field due to a charged rod at a point on its perpendicular bisector? That is of relevance here.

Because $$\vec{E}_{perp}=\frac{kq}{r^2}$$, and at the perpendicular bisector $$r=\sqrt{\left(\frac{d} {2}\right)^2+d^2} = \frac{9\sqrt{3}}{2}$$, the expresion is then $$dqdx\cdot\frac{k}{\frac{9\sqrt{3}}{2}^2}$$. Though I am not sure I fully understand what you mean, since I'm unsure why it matters what they are at the perpendicular bisector if I have a general formula for any point on the line.

 

[kuruman]

and at the perpendicular bisector
$$r=\sqrt{\left(\frac{d} {2}\right)^2+d^2} = \frac{9\sqrt{3}}{2}$$

If ##d## is the side of the triangle, then what is distance ##r## given in the above the expression? Not all points on the triangle are at the same distance from the center therefore they contribute differently to the net field at the center.

You need to rethink your solution. I suggest that you read and understand the derivation under "Electric Field of Line Charge" here http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html. You need to add three such contributions, one from each side, to the total. Note that the electric field is a vector so you need to add the three fields as vectors. Can you just look at the problem and figure out the direction of the net electric field? Hint: Think symmetry.

 

[Roverse]

If ##d## is the side of the triangle, then what is distance ##r## given in the above the expression? Not all points on the triangle are at the same distance from the center therefore they contribute differently to the net field at the center.

You need to rethink your solution. I suggest that you read and understand the derivation under "Electric Field of Line Charge" here http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html. You need to add three such contributions, one from each side, to the total. Note that the electric field is a vector so you need to add the three fields as vectors. Can you just look at the problem and figure out the direction of the net electric field? Hint: Think symmetry.

Ah yeah that was a typo (out of confusion, I thought you wanted the value if a point charge was at the bisector). In my above equation. the r is actually $$\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+\left(d-\left(\frac{d}{2}\right)^2\right)}$$. where x goes from 0 to 18cm along the rod. I have the vectors drawn in the picture, no?

Oh an r is the distance to the center (distance from point charge) (see figure).How is my line charge integral incorrect (I'm pretty sure I understand the line charge derivation)? I have the Total/Length * tiny distance * k / changing distance * sin(theta) as o/h.

1. found one mistake so far. It is where I say the distance to the center is (d + (d/2)^2)^1/2 and not that over 2.

 

[kuruman]

You still have not posted an expression for the electric field at the perpendicular bisector of a rod of length ##d##. The expression must have the given quantities, in this case the total charge on the rod ##q##, the side of the triangle ##d## and of course the constant ##k##, nothing else. Their values alone determine the magnitude of the field. If you write it in terms of something like ##x## "where x goes from 0 to 18cm along the rod" as you say, what specific value are you going to use for ##x##? You need to do an integral over x as the reference I gave you suggests.

I have the vectors drawn in the picture, no?

Yes, you have. What are their magnitudes?

 

[Roverse]

You still have not posted an expression for the electric field at the perpendicular bisector of a rod of length ##d##. The expression must have the given quantities, in this case the total charge on the rod ##q##, the side of the triangle ##d## and of course the constant ##k##, nothing else. Their values alone determine the magnitude of the field. If you write it in terms of something like ##x## "where x goes from 0 to 18cm along the rod" as you say, what specific value are you going to use for ##x##? You need to do an integral over x as the reference I gave you suggests.

Yes, you have. What are their magnitudes?

I understand the way the reference did it. I actually just found the error (I forgot to divide by a factor of 2 (for my little half distance across the height, where before I had the entire). My answer is correct now. Thanks for the help!

 

[Roverse]

Quick correction: I didn't correctly write the solution integral on here even though I did in my own calculations:

$$\int_0^d\frac{\left(k\cdot\frac{charge}{\operatorname{length}}\cdot\left(\frac{\operatorname{length}}{2}\right)\cdot\left(\tan\left(30degrees\right)\right)\right)}{\left(\left(\left(\frac{\operatorname{length}}{2}\right)\cdot\tan\left(30degrees\right)^2+\left(\frac{\operatorname{length}}{2}-x\right)^2\right)\right)^{\frac{3}{2}}}dx$$

 

[kuruman]

That is correct except that in the denominator the radical should not be there.

 

[Roverse]

That is correct except that in the denominator the radical should not be there.

Yeah I'm still learning latex, misplaced a curly brace. One moment.

Is it good now?

 

[kuruman]

Is it good now?

I don't know because I can't see it. While you're at it, can you make the equation more compact and thus more legible? For example, ##length## can be replaced with ##d## and ##30degrees## with ##30^o## (30^o).

On Edit: Did you edit the equation in post #8? If so it's still incorrect. The outer two pairs of parentheses in the denominator can be replaced with a single pair and it will make no difference to what the expression is saying.