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...

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?

Asked on

1 Answer | Add Yours

pramodpandey's profile pic

Posted on

`+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.

 

We’ve answered 333,587 questions. We can answer yours, too.

Ask a question