Answer to Question #182545 in Electric Circuits for farouk

Question

Q1 = 7.5 uC, Q2= 7.2 uc , Q3= -12.9uC ,

The component of the resultant electric force on Q3 on the X-axis is

Answer 1

To be given in question

Q₁₌7.5$\mu c$

Q₂₌7.2$\mu c$

Q₃₌-12.9$\mu c$

To be asked in question

Electric force on x-axis

We know that

$F_{13}=k\frac {q_{1}q_{3}}{r^2}$ $\rightarrow(1)$

$F_{23}=k\frac {q_{2}q_{3}}{r^2} \rightarrow(2)$

$F=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos\theta}$

$\theta=90°$

Put $\theta$ value

$F=\sqrt{F_{13}^2+F_{23}^2}\rightarrow(3)$

equation (1)put values

$F_{13}=-\frac{9\times10^9\times7.5\times10^{-6}\times12.9\times10^{-6}}{r_{13}^2}\rightarrow(4)$

$F_{13}=-\frac{0.87075}{r_{13}^2}\rightarrow(4)$

$F_{23}=-\frac{9\times10^9\times7.2\times10^{-6}\times12.9\times10^{-6}}{r_{23}^2}$

$F_{23}=\frac{-0.83592}{r_{23}^2}\rightarrow(5)$

equation (3); equation (4)and eqution (5) we can written as

$F=\sqrt(-\frac{0.87075}{r_{13}^2})^2+(\frac{-0.83592}{r_{23}^2})^2\rightarrow(6)$

equation (6)put value r₁₃and r₂₃

and find the resultant force