Answer to Question #121823 in Electricity and Magnetism for maggie

Question

Three charged particles are arranged on corners of square .suppose the +2q charge at the left corner is removed.then does the magnitude of the field at the upper right corner

Answer 1

$\overrightarrow{E_1}+\overrightarrow{E_2}$

where $E_1$ is electric field due to charge 1 and $E_2$ is electric filed due to charge 2

and since $E_1\ \&\ E_2$ are perpendicular hence the resultant magnitude will be $\sqrt{(E_1)^2+(E_2)^2}$

$E_1=\frac{k2q}{a^2}$

$E_2=\frac{k2q}{a^2}$

resultant electric field = $\dfrac{2\sqrt2kq}{a^2}$