Question

Two point charges $q_1$ and $q_2$ are placed at $(0,0,0)$ and $(1,2,2)m$ respectively. They repel each other with force of $3N$. The force on $q_2$ due to $q_1$ is $F_{21}=(x\hat{i}+y\hat{j}+z\hat{k})N$. The value of $x+y+z$ is ___

Answer 1

Answer: 5

$F_{21}=F_{21}\hat{e}=F_{21}\frac{AB}{|AB|}=F_{21}\frac{(\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{1+4+4}}$
$=\frac{3}{3}(\hat{i}+2\hat{j}+2\hat{k})=\hat{i}+2\hat{j}+2\hat{k}$
$\therefore x=1,y=2,z=2$
$\therefore x+y+z=5$