Electric field at the center of the equilateral triangle

[hellowmad]

Homework Statement:

Consider an equilateral triangle of side 20. cm. A charge of +2.0 μC is placed at one vertex and charges of -4.0 μC are placed at the other two vertices. Determine the magnitude and direction of the electric field at the center of the triangle.

Relevant Equations:

E = kQ1/r^2

I've found the distance from each point to the center, which is equal to r=20x1.732/3 = 11.55 cm.
I find out that E2 and E3 due to -4µEyC on x-direction canceled each other.
The E2y = E3Y = EY = E2Ycos60 = E2/2 = [(KQ2)/r^2]/2
So the net E-field:
E = E1 +E2y+E3Y
=kQ1/r^2 + [(KQ2)/r^2]/2 + [(KQ2)/r^2]/2
= kQ1/r^2 + (KQ2)/r^2
= (k/r^2) (|Q1|+|Q2|)
= (9 x 10^9/0.1155^2) [(0.000002) + (0.000006)]
= 4.05x10^ N/C
= 4.05x10^ V/m

However, answer key say it should be 2.7x10^6 V/m. Do I do something wrong?

 

[haruspex]

kQ1/r^2 + (KQ2)/r^2
= (k/r^2) (|Q1|+|Q2|)

Where did the modulus signs come from?

 

[hellowmad]

It is not modulus sign. it just means absolute number only. So, it stands as 2µC and 4µC.

 

[haruspex]

It is not modulus sign. it just means absolute number only. So, it stands as 2µC and 4µC.

That is what I mean by modulus here, the absolute value. Why are you taking the absolute values of the charges?

 

[hellowmad]

No matter what charge places on the center of triangle, the E field direction of two -4µC's Y-direction is the same as that of 2µC one. The final E-field is the sum of them, So I take the absolute values of the charges here. Thanks

 

[TSny]

= (k/r^2) (|Q1|+|Q2|)
= (9 x 10^9/0.1155^2) [(0.000002) + (0.000006)]

|Q2| is 4 μC , not 6 μC. But I think this is just a typo. The total inside the square brackets is 6 μC.

= 4.05x10^ N/C
= 4.05x10^ V/m

This looks correct to me if you intended the power of 10 to be 6.

However, answer key say it should be 2.7x10^6 V/m.

I think the key is wrong and your answer is right.

 

[hellowmad]

Thank TSny. Yes couple typo. Sorry for that. My answer is 4.05x10^6 V/m.

Thank you everyone

 

[haruspex]

No matter what charge places on the center of triangle, the E field direction of two -4µC's Y-direction is the same as that of 2µC one. The final E-field is the sum of them, So I take the absolute values of the charges here. Thanks

Right, but that's a dodgy way to fix it. Your working should have been

E = E1 +E2y+E3Y
=kQ1/r^2 - [(KQ2)/r^2]/2 - [(KQ2)/r^2]/2
= kQ1/r^2 - (KQ2)/r^2
= (k/r^2) (Q1-Q2)

The minus signs in line 2 reflect the fact that those charges are on the opposite side.

Btw, the book answer is exactly 2/3 of yours. That should be a clue as to where they went wrong, but I can't pick it.

 

[hellowmad]

Yes, this is the point I post here to see if I forget something or make a mistake.

 

[kuruman]

Yes, this is the point I post here to see if I forget something or make a mistake.

I repeated the calculation and I agree with your answer. It does not look like you have made a mistake or overlooked something.

 

[hellowmad]

Thank you so much kuruman