Answer to Question #174420 in Electric Circuits for JC YOUNG

Question

Calculate the net electric field at point A for the following charge distribution. 

q1 northern most point charge of -5x10^-5

q2 Eastern most point charge of -3.0x10^-5

point a the western most point

point a to q1 is 5.0m

point a to q2 is 5.0m

Answer 1

Answer

Electric field due to q$_1$ On A

$E_1=\frac{kq_1}{r_1^2}\\=\frac{9\times10^9\times5\times10^{-5}}{5^2}\\=1.8\times10^4 V/m$

With direction

$E_1=1.27\times10^4 (-i-j)V/m$

Electric field due to q₂ A

$E_2=\frac{kq_2}{r_2^2}\\=\frac{9\times10^9\times3\times10^{-5}}{5^2}\\=0.72\times10^4V/m$

With direction

$E_2=0.72\times10^4(-j)V/m$

So net electric field On Point A

$E=E_1+E_2$

=$1.27\times10^4 (-i) +1.99\times10^4( -j)V/m\\=-[1.27i+1.99j]\times10^4 V/m$