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Q.

Two point charges q1 and q2 are placed at (0,0,0) and (1,2,2) m respectively. They repel each other with force of 3 N. The force on q2 due to q1 is F21=(xi^+yj^+zk^)N The value of x+y+z is _________.

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answer is 5.

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Detailed Solution

F21=F21e^=F21AB|AB|=F21(i^+2j^+2k^)1+4+4 =33(i^+2j^+2k^)=i^+2j^+2k^∴ x=1,y=2,z=2∴ x+y+z=5
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