A charge +2 `muC` is placed at a point A (2,1,0). Another charge `-3muC` is placed at a point B (4,2,-1). Calculate net force on a charge `+5 muC` placed at a point C (-1,3,2).

Explanation

If `vecr` represents a vector joining the two charges along the direction of force. Then the force in vector form can be written as follows :
`vecF=(Qq)/(4piepsilon_0)[vecr/r^3]`
Notice that is the given figure the points are not shown on appropriate locations as per their coordinates . But the fact is that you do not require this , as you can work just on the basis of given coordinates . Here , you should note that the direction of `vecr_1` and `vecr_2` are selected as per the direction of forces `vecF_1` and `vecF_2` , respectively.

`vecr_1=(-1-2)hati+(3-1)hatj+(2-0)hatk=-3hati+2hatj+2hatk`
`vecr_2=(4+1)hati+(2-3)hatj+(-1-2)hatk=5hati-hatj-3hatk`
`r_1=sqrt((-3)^2+(2)^2+(2)^2)=sqrt17`
`r_2=sqrt((5)^2+(-1)^2+(-3)^2)=sqrt35`
`vecF_1=(9xx10^9)(2xx10^(-6))(5xx10^(-6))vecr_1/r_1^3 =0.09 vecr_1/r_1^3`
`vecF_2=(9xx10^9)(3xx10^(-6))(5xx10^(-6))vecr_2/r_2^3=0.135 vecr_2/r_2^3`
Now substituting the values
`vecF_1=0.09vecr_1/r_1^3=0.09/17^(3//2) (-3hati+2hatj+2hatk)`
`=(0.00130)(-3hati+2hatj+2hatk)`
`vecF_2=0.135 vecr_2/r_2^3 =0.135/35^(3//2)(5hati-hatj-3hatk)`
`=(0.00065)(5hati-hatj-3hatk)`
On solving we get
`vecF=vecF_1+vecF_2=0.00065(-hati+3hatj+hatk)`
Magnitude of net force F=0.00065 `(sqrt11)`
=0.0022 newton