Question

Two charges Q1 of 3 C and Q2 of 5 C are located at ri a) Find the vector force F12 on Qs due to Q2 in terms of unit vectors I, j, and k b) Extract the x, y and z components of force c) Find the unit vector in the direction of the force F12. d) Find the direction angles 21 + 4 j + 3k and r,-31+5-6k .

Answer 1

a)

$\underset{F_{12}}{\rightarrow}$ = force on charge Q₁ by charge Q₂ = k Q₁ Q₂ ($\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ ) /(|$\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ |)³

$\underset{r_{1}}{\rightarrow}$ = 2 i + 4 j + 3 k

$\underset{r_{2}}{\rightarrow}$= 3 i + 5 j - 6 k

$\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ = (2 i + 4 j + 3 k) - (3 i + 5 j - 6 k ) = - i - j + 9 k

|$\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ | = sqrt((-1)² + (-1)² + (9)²) = 9.11

$\underset{F_{12}}{\rightarrow}$ = k Q₁ Q₂ ($\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ ) /(|$\underset{r_{1}}{\rightarrow}$- $\underset{r_{2}}{\rightarrow}$ |)³

$\underset{F_{12}}{\rightarrow}$ = (9 x 10⁹) (3) (5) (- i - j + 9 k ) /(9.11)³

$\underset{F_{12}}{\rightarrow}$ = (1.8 x 10⁸) (- i - j + 9 k )

$\underset{F_{12}}{\rightarrow}$ = - (1.8 x 10⁸) i - (1.8 x 10⁸) j + (16.2 x 10⁸) k

b)

x- component : - (1.8 x 10⁸)

y-component : - (1.8 x 10⁸)

z-component : (16.2 x 10⁸)

c)

|$\underset{F_{12}}{\rightarrow}$| = sqrt((- (1.8 x 10⁸) )² + (- (1.8 x 10⁸) )² + (16.2 x 10⁸)²) = 1.6 x 10⁹

unit vector is given as

unit vector = $\underset{F_{12}}{\rightarrow}$ / |$\underset{F_{12}}{\rightarrow}$| = (- (1.8 x 10⁸) i - (1.8 x 10⁸) j + (16.2 x 10⁸) k) / (1.6 x 10⁹ )