Question

Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.

Answer 1

We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point

`E=\sum_{n\mathop{=}0}^(\infty)E_i`

To calculate the distance R we have a triangle

`R=\sqrt{(2a^2)/(4)}=(a)/(√(2))=0.014m=1.41cm`
`\begin{gathered} \sum_{n\mathop{=}0}^(\infty)Ex=(9\cdot10^9)/(2\cdot10^(-4))\cdot(cos45)\cdot(6C)=1.91\cdot10^(14)N/C \\ \sum_{n\mathop{=}0}^(\infty)Ey=4.5\cdot10^(13)\cdot(cos45\degree)(2C)=0.636\cdot10^(14)N/C \\ Etot=√(Ex^2+Ey^2)=2.01\cdot10^(14)N/C \end{gathered}`

Is important to remember that E has direction so you have to calculate each axis, x, and y