Answer to Question #209546 in Electric Circuits for John Mark

Question

A charge q1= 7.00 µC is located at the origin, and a second charge q2 = -5.00 µC is located on the x axis, 0.30 m from the origin. (A) Find the electric field at the point P, which has coordinates (0, 0.400) m. (B) If a charge q3 = -3.00 µC is placed at point P, what is the coulomb force experienced by this charge due to other charges q1 and q2. 

Answer 1

Gives

$q_1=7.00\mu C;q_2=-5.00\mu C$

$$r_1=0.40m;r_2=\sqrt{.40^2+.30^2}=0.50m$$

$sin\theta=\frac{0.40}{0.50}=\frac{4}{5}$

$cos\theta=\frac{0.30}{0.40}=\frac{3}{4}$

$$E_1=\frac{kq_1}{r_1^2}=\frac{{9\times10^9}\times7\times10^{-6}}{0.40^2}=3.9310^{5}N/C$$

$$E_2=\frac{kq_2}{r_2^2}=-\frac{{9\times10^9}\times5\times10^{-6}}{0.50^2}=1.8\times10^{5}N/C$$

$E_2$ x-axis components

$E_2cos\theta$

$E_2$ y-axis components

$-E_2sin\theta$

$E_1=3.93\times10^5\hat{j}N/C$

$$E_2=1.08\hat{i}+{(-1.44)}\hat{j}\times10^{5}N/C$$

$E=E_1+E_2$

$$E=1.1\times10^5\hat{i}+2.5\times10^{5}\hat{j} N/C$$

Force

$q_3=-3.0\mu C$

$F_{13}=\frac{Kq_1q_3}{r_2^2}$

$$F_{13}=-\frac{{9\times10^9}\times7\times10^{-6}\times3\times10^{-6}}{0.50^2}=0.756N$$

$F_{23}=\frac{Kq_2q_3}{r_1^2}$

$$F_{23}=\frac{{9\times10^9}\times5\times10^{-6}\times3\times10^{-6}}{0.40^2}=0.843N$$