Net Electric Field at the center of a square

[kemcco1955]

Homework Statement

What net electric field do the particles of problem #1 produce at the square's center? Problem number one has a square with edge lengths of 24.00cm and charges of q1==8.0e-6C, q2=-8.0e-6C, q3=-8.0e-6, and q4=+8.0. He wants the answer for the electric field in component form(Etotal= N/C x + N/Cy. I do not know how to find electric field when using different components.

Homework Equations

Ecenter=kq/r^2

The Attempt at a Solution

I know that q1 and q3 cancel/ equal zero because they are both negative and on the diagonal from one another. The vector for q2 points away from the charge and the vector for q4 points toward the charge. I find E2 using (8.99e9)(-8e-6)/.1697^2. And for E4 I get the same but with an opposite sign. The r it got to be .1697 m by taking the sqrt of .24^2+.24^2 =.3394/2=.1697. The angel, I think is 45 degrees. I am having difficulty finding how to get the E total into components I know that you can take an answer and multiply it by cos45 and sin45(which are the same), but I don't know how to set up the vectors to solve...I hope that makes some since...I am very confused.

 

[alphysicist]

Hi kemcco1955,

Homework Statement

What net electric field do the particles of problem #1 produce at the square's center? Problem number one has a square with edge lengths of 24.00cm and charges of q1==8.0e-6C, q2=-8.0e-6C, q3=-8.0e-6, and q4=+8.0. He wants the answer for the electric field in component form(Etotal= N/C x + N/Cy. I do not know how to find electric field when using different components.

Homework Equations

Ecenter=kq/r^2

The Attempt at a Solution

I know that q1 and q3 cancel/ equal zero because they are both negative and on the diagonal from one another. The vector for q2 points away from the charge and the vector for q4 points toward the charge.

No, q2 is a negative charge, so its field points toward itself. q4 is a positive charge so its field points away from itself.

I find E2 using (8.99e9)(-8e-6)/.1697^2. And for E4 I get the same but with an opposite sign. The r it got to be .1697 m by taking the sqrt of .24^2+.24^2 =.3394/2=.1697. The angel, I think is 45 degrees. I am having difficulty finding how to get the E total into components I know that you can take an answer and multiply it by cos45 and sin45(which are the same), but I don't know how to set up the vectors to solve...I hope that makes some since...I am very confused.

Draw the E total vector with the magnitude labelled and with the correct direction. Then it's just a matter of breaking it into x and y components like any other vector. For example,

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec5

Of course, depending on which direction the E-field is pointing, you might get one or both components to be negative.