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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 ata point C (-1,3,2). |
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Answer» Solution :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 |
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