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Three point charges +6.4e-6, +2.5e-6, and -3.6e-6 lie along the x axis at 0m, .015m and...

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sae755 | eNotes Newbie

Posted April 22, 2013 at 5:20 PM via web

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Three point charges +6.4e-6, +2.5e-6, and -3.6e-6 lie along the x axis at 0m, .015m and .055m. What is the force exerted on q2 by the other two charges? What is the force exerted on q3 by the other two charges?

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted April 23, 2013 at 7:23 AM (Answer #1)

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`+6.4xx10^(-6)`         `+2.5xx10^(-6)`                    `-3.6xx10^(-6)`
 
`Q_1` ______`r_1` _________`Q_2`________`r_2`________  `Q_3`
             .015m                                .055m

Let the above line be a positive x-axis and
 
              `r_1`=0.015m, `r_2`=0.055m
(i) Force exerted on `Q_2` by the other two charges is the sum of the force of repulsion by `Q_1` and the force of attraction by`Q_3`.
`F_(12)=(kQ_1Q_2)/(r_1)^2`
 
`F_(12)=(8.99xx10^9xx6.4xx10^(-6)xx2.5xx10^(-6))/(1.5xx10^(-2))^2`
```=6.392xx10^2 N`   to  the right
`F_32=(kQ_3Q_2)/(r_2)^2`
`=(8.99xx10^9xx3.6xx10^(-6)xx2.5xx10^(-6))/(5.5xx10^(-2))^2`
`=2.67xx10^1 N`  to the right
 
Resultant force on`Q_2=F_12 +F_32` 
 
`=639.2+26.7=665.9 N` 

(ii)Force exerted on `Q_3`  by the other two charges is the sum of the force of attraction by `Q_1` & `Q_3.`
``
`F_13=(kQ_1Q_3)/(r_1+r_2)^2`
`=(8.99xx10^9xx6.4xx10^(-6)xx3.6xx10^(-6))/(7xx10^(-2))^2`
`=4.22xx10^1 N`to the left
 
Resultant force on`Q_3=F_13+F_23` 
 
`=-42.2-26.7`
 
`=--68.9`  N to the left.

 

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