darioslc
- 9
- 2
- 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
$$\vec{E}(0,0)=-\frac{\sqrt{2}Qk}{d^2}\hat{x}-\frac{\sqrt{2}Qk}{d^2}\hat{y}$$
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