Answer to Question #126469 in Electric Circuits for thankssss

Question

Two point charges form a right-angled triangle with the point A at the origin. Their charges are Q₂ = 6 x 10^-9 C = 6 nC and Q₃ = -3 x 10^-9 C = -3 nC. The distance between point A and Q₂ is 5 x10^-2 m and the distance between point A and Q₃ is 3 x 10^-2 m. What is the net electric field measured at A from the two charges?

Answer 1

As per the question,

The given charges are $Q_2=6\times 10^{-9}C$

Distance between A and $Q_2$ $(r_2)=5\times 10^{-2}m$

$Q_3=-3\times 10^{-9}C$

Distance between A and $Q_3$ $(r_3)=3\times 10^{-2}m$

A is at the origin, let $Q_2$ is on the x axis and $Q_3$ is on the y-axis.

Hence,$E_i=\frac{Q_2}{4\pi \epsilon_o r_2^2}\hat{i}=9\times 10^9\times\frac{6\times 10^{-9}C}{(5\times 10^{-2}m)^2}\hat{(-i)}$

$=\frac{-54}{25\times 10^{-4}}\hat{i}=-2.16\times 10^4\hat{i}$

$E_j=\frac{Q_3}{4\pi \epsilon_o r_3^2}\hat{j}=9\times 10^9\times\frac{-3\times 10^{-9}C}{(3\times 10^{-2}m)^2}\hat{(-j)}$

$=3\times 10^4\hat{j}$

$E_{net}=-2.16\times 10^4\hat{i}+3\times 10^4\hat{j}$

$|E_{net}|=\sqrt{E_i^2+E_j^2} =3.7\times 10^4 N/C$