- #1

hellowmad

- 11

- 2

- 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?

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?