tom75
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I have just begun studying electrostatic and I'm trying to do this exercize:
We have a square with charges +q , -2q, +2q, -q1)Compute the electrostatic field \vec{E}at the center of the square.
I did this way :
I find \vec{E_A}=\frac{q}{2 \pi \epsilon_0} \vec{u}
{E_B}=\frac{-q}{ \pi \epsilon_0} \vec{u}
{E_C}=\frac{q}{ \pi \epsilon_0} \vec{u}
{E_D}=\frac{-q}{2 \pi \epsilon_0} \vec{u}
Then with projection :
E_A=\frac{q}{2 \pi \epsilon_0}*cos(45)=\frac{\sqrt{2}q}{4\pi \epsilon_0}
E_B=\frac{-q}{ \pi \epsilon_0}*cos(45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}
E_C=\frac{q}{2 \pi \epsilon_0}*sin(-45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}
E_D=\frac{-q}{2 \pi \epsilon_0}*sin(45)=\frac{-\sqrt{2}q}{4\pi \epsilon_0}
Finally E_{total}=\frac{-\sqrt{2}q}{\pi \epsilon_0}
Is-it correct ? I'm not sure of my way of reasoning and the projection.
Thank you
We have a square with charges +q , -2q, +2q, -q1)Compute the electrostatic field \vec{E}at the center of the square.
I did this way :
I find \vec{E_A}=\frac{q}{2 \pi \epsilon_0} \vec{u}
{E_B}=\frac{-q}{ \pi \epsilon_0} \vec{u}
{E_C}=\frac{q}{ \pi \epsilon_0} \vec{u}
{E_D}=\frac{-q}{2 \pi \epsilon_0} \vec{u}
Then with projection :
E_A=\frac{q}{2 \pi \epsilon_0}*cos(45)=\frac{\sqrt{2}q}{4\pi \epsilon_0}
E_B=\frac{-q}{ \pi \epsilon_0}*cos(45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}
E_C=\frac{q}{2 \pi \epsilon_0}*sin(-45)=\frac{-\sqrt{2}q}{2\pi \epsilon_0}
E_D=\frac{-q}{2 \pi \epsilon_0}*sin(45)=\frac{-\sqrt{2}q}{4\pi \epsilon_0}
Finally E_{total}=\frac{-\sqrt{2}q}{\pi \epsilon_0}
Is-it correct ? I'm not sure of my way of reasoning and the projection.
Thank you