# 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?

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Student Comments

pramodpandey | Student

`+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`

(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.`

(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.