Electric field created by point charges

[darioslc]

Homework Statement

Find the electric field in the center of square, if in his three vertices there are three puntual charges:
2q, 2q and Q and the side is d:

2q *-------------
| |
| |
Q *-------------* 2q

Relevant Equations

The only relevant equation is $$E=\frac{kq}{r^2}$$

Hello, I reasoned by simmetry, the two charges with value 2q not contributed at field because there are equidistant at point and are similar charges. Therefor only survival the field due to Q, using the definition for electric field of the puntual charge:
$$\vec{E}(0,0)=-\frac{\sqrt{2}Qk}{d^2}\hat{x}-\frac{\sqrt{2}Qk}{d^2}\hat{y}$$
because I set the square in the center, then:
\begin{align}
E(x,y)=&-\frac{Qk}{[(x+d/2)^2+(y+d/2)^2]^{3/2}}(x+d/2,y+d/2)\text{ evaluated in the (0,0)}\\
E(0,0)=&-\frac{Qk}{[(d/2)^2+(d/2)^2]^{3/2}}(d/2,d/2)\nonumber\\
E(0,0)=&-\frac{Qk}{[2(d/2)^2]^{3/2}}(d/2,d/2)\nonumber\\
E(0,0)=&-\frac{Qk}{[d^2/2]^{3/2}}(d/2,d/2)\nonumber\\
E(0,0)=&-\frac{\sqrt{2}Qk}{d^2}(1,1)\nonumber
\end{align}
which makes senses, because Q is negative.

¿That is correct? because I doubt a little bit

 

[anuttarasammyak]

When we put charge Q on the absent corner, center E is obviously zero.

The configuration of single charge -Q as shown below should give same center E we want.

This is my estimation. Does it show same result as yours ?

 

[kuruman]

I agree that the contributions of the two charges $q$ cancel each other. This leaves the contribution from charge $Q$ at the lower left corner. The general expression for the electric field at position $\mathbf r$ due to charge $Q$ at position $\mathbf r'$ is $$\mathbf E=kQ\frac{(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}.$$ If you put the origin of coordinates at the center of the square, $\mathbf r=0$. What is $\mathbf r'$ in this case? What do you get when you substitute in the general expression?

 

[darioslc]

When we put charge Q on the absent corner, center E is obviously zero.
View attachment 351358
The configuration of single charge -Q as shown below should give same center E we want.
View attachment 351359
This is my estimation. Does it show same result as yours ?

But when you put the charge -Q in the opposite corner (when in original scheme it's absent charge), the field is the same but not the sense, no?
The scheme with the center (0,0) is:
2q *-----------------
| |
| (0,0) |
| |
-Q *----------------* 2q

Excuse me, I see that the sign of Q is -Q, not +Q like in the initial. Anyway, the field value is correct using -Q.
Now I see your reasoning.

Then finally the field due both 2q is zero in the center, except for -Q

 

[kuruman]

But when you put the charge -Q in the opposite corner (when in original scheme it's absent charge), the field is the same but not the sense, no?

No. Charge $+Q$ at the lower left corner produces exactly the same field (magnitude and direction) as charge $-Q$ at the upper right corner.

 

[Orodruin]

No. Charge $+Q$ at the lower left corner produces exactly the same field (magnitude and direction) as charge $-Q$ at the upper right corner.

… at the center of the square.

May seem like nitpicking but an important point to note.

 

[kuruman]

… at the center of the square.

May seem like nitpicking but an important point to note.

Of course.