# Electric field at the center of the equilateral triangle

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?

Homework Helper
Gold Member
2022 Award
kQ1/r^2 + (KQ2)/r^2
= (k/r^2) (|Q1|+|Q2|)
Where did the modulus signs come from?

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

Homework Helper
Gold Member
2022 Award
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?

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

Homework Helper
Gold Member
= (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.

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

Thank you everyone

Homework Helper
Gold Member
2022 Award
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.

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

Homework Helper
Gold Member
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.

• • 