Question #182545

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


1
Expert's answer
2021-04-19T17:44:40-0400

To be given in question

Q1=7.5μc\mu c

Q2=7.2μc\mu c

Q3=-12.9μc\mu c

To be asked in question

Electric force on x-axis

We know that

F13=kq1q3r2F_{13}=k\frac {q_{1}q_{3}}{r^2} (1)\rightarrow(1)

F23=kq2q3r2(2)F_{23}=k\frac {q_{2}q_{3}}{r^2} \rightarrow(2)

F=F132+F232+2F13F23cosθF=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos\theta}

θ=90°\theta=90°

Put θ\theta value

F=F132+F232(3)F=\sqrt{F_{13}^2+F_{23}^2}\rightarrow(3)

equation (1)put values

F13=9×109×7.5×106×12.9×106r132(4)F_{13}=-\frac{9\times10^9\times7.5\times10^{-6}\times12.9\times10^{-6}}{r_{13}^2}\rightarrow(4)

F13=0.87075r132(4)F_{13}=-\frac{0.87075}{r_{13}^2}\rightarrow(4)

F23=9×109×7.2×106×12.9×106r232F_{23}=-\frac{9\times10^9\times7.2\times10^{-6}\times12.9\times10^{-6}}{r_{23}^2}

F23=0.83592r232(5)F_{23}=\frac{-0.83592}{r_{23}^2}\rightarrow(5)

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

F=(0.87075r132)2+(0.83592r232)2(6)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



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