To be given in question
Q1= 7.5μ c \mu c μ c
Q2= 7.2μ c \mu c μ c
Q3= -12.9μ c \mu c μ c
To be asked in question
Electric force on x-axis
We know that
F 13 = k q 1 q 3 r 2 F_{13}=k\frac {q_{1}q_{3}}{r^2} F 13 = k r 2 q 1 q 3 → ( 1 ) \rightarrow(1) → ( 1 )
F 23 = k q 2 q 3 r 2 → ( 2 ) F_{23}=k\frac {q_{2}q_{3}}{r^2} \rightarrow(2) F 23 = k r 2 q 2 q 3 → ( 2 )
F = F 13 2 + F 23 2 + 2 F 13 F 23 c o s θ F=\sqrt{F_{13}^2+F_{23}^2+2F_{13}F_{23}cos\theta} F = F 13 2 + F 23 2 + 2 F 13 F 23 cos θ
θ = 90 ° \theta=90° θ = 90°
Put θ \theta θ value
F = F 13 2 + F 23 2 → ( 3 ) F=\sqrt{F_{13}^2+F_{23}^2}\rightarrow(3) F = F 13 2 + F 23 2 → ( 3 )
equation (1)put values
F 13 = − 9 × 1 0 9 × 7.5 × 1 0 − 6 × 12.9 × 1 0 − 6 r 13 2 → ( 4 ) F_{13}=-\frac{9\times10^9\times7.5\times10^{-6}\times12.9\times10^{-6}}{r_{13}^2}\rightarrow(4) F 13 = − r 13 2 9 × 1 0 9 × 7.5 × 1 0 − 6 × 12.9 × 1 0 − 6 → ( 4 )
F 13 = − 0.87075 r 13 2 → ( 4 ) F_{13}=-\frac{0.87075}{r_{13}^2}\rightarrow(4) F 13 = − r 13 2 0.87075 → ( 4 )
F 23 = − 9 × 1 0 9 × 7.2 × 1 0 − 6 × 12.9 × 1 0 − 6 r 23 2 F_{23}=-\frac{9\times10^9\times7.2\times10^{-6}\times12.9\times10^{-6}}{r_{23}^2} F 23 = − r 23 2 9 × 1 0 9 × 7.2 × 1 0 − 6 × 12.9 × 1 0 − 6
F 23 = − 0.83592 r 23 2 → ( 5 ) F_{23}=\frac{-0.83592}{r_{23}^2}\rightarrow(5) F 23 = r 23 2 − 0.83592 → ( 5 )
equation (3); equation (4)and eqution (5) we can written as
F = ( − 0.87075 r 13 2 ) 2 + ( − 0.83592 r 23 2 ) 2 → ( 6 ) F=\sqrt(-\frac{0.87075}{r_{13}^2})^2+(\frac{-0.83592}{r_{23}^2})^2\rightarrow(6) F = ( − r 13 2 0.87075 ) 2 + ( r 23 2 − 0.83592 ) 2 → ( 6 )
equation (6)put value r13 and r23
and find the resultant force
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