# Electric field at the center of the equilateral triangle

In summary, In the above conversation, the distance from each point to the center is equal to r=20x1.732/3 = 11.55 cm. E2 and E3 due to -4µEyC on x-direction canceled each other, resulting in the net E-field being E = E1 +E2y+E3Y. The answer key says the net E-field should be 2.7x10^6 V/m, but kQ1/r^2 + (KQ2)/r^2 = (k/r^2) (Q1-Q2) results in the final E-field being 4.05x10^ N/
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?

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.

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

= (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

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.

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.

Thank you so much kuruman

kuruman

## What is an electric field?

An electric field is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

## What is the center of an equilateral triangle?

The center of an equilateral triangle is the point that is equidistant from all three vertices of the triangle. It is the point of symmetry for the triangle.

## What is the electric field at the center of an equilateral triangle?

The electric field at the center of an equilateral triangle is zero. This is because the electric fields created by each of the three charges at the vertices cancel each other out due to the symmetry of the triangle.

## How is the electric field at the center of an equilateral triangle calculated?

The electric field at the center of an equilateral triangle can be calculated using Coulomb's law, which states that the electric field at a point is equal to the sum of the electric fields created by each individual charge at that point. In this case, the electric fields created by the three charges at the vertices can be added together to find the total electric field at the center.

## What is the significance of the electric field at the center of an equilateral triangle?

The fact that the electric field at the center of an equilateral triangle is zero has practical applications in fields such as engineering and physics. It can be used to design and analyze systems that involve equilateral triangles, such as antennas and electronic circuits. It also helps to understand the concept of symmetry in electric fields and how it can affect the overall behavior of a system.

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