Question

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 3.75×10^-6 C charges ( E = ?) . And what is the direction of the electric field at the corner ?

Answer 1

The electric field E, at the corner of the square is the sum of the fields due to the other 3 charges.

k=8.99x10^9

$E_1=E_{1x}=k\frac{Q}{r^2}$ because $E_{1y}=0$

$E_2=k\frac{Q}{2r^2}$

$E_{2x}=k\frac{Q}{2r^2}cos(45)$ Note: 2r^2 (Pythagorian theorem): The distance to the point in consideration is the hypotenuse of a triangle with a value of adjacent and opposite ---> r

$E_{2y}=k\frac{Q}{2r^2}sin(45)$

$E_{3}=E_{3y}=k\frac{Q}{r^2}$   because $E_{3x}=0$

$E_{total}x=E_{1x}+E_{2x}+E_{3x}=E_{1x}+E_{2x}$

$E_{total}x=$$E_{total}x=8.99x10^9*\frac{3.75x10^{-6}}{2.42^2}+8.99x10^9*\frac{3.75x10^{-6}cos(45)}{2*2.42^2}=7,791.76[N/C]$

$E_{total}y=8.99x10^9*\frac{3.75x10^{-6}}{2.42^2}+8.99x10^9*\frac{3.75x10^{-6}sin(45)}{2*2.42^2}=7,791.76[N/C]$

$E_total=\sqrt{E^2_{totaly}+E^2_{totalx}}=2,035.24N/C$