First slide

Electrostatic force

Question

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 q₁is $F_{21} = (x \hat{i} + y \hat{j} + z \hat{k}) N$ The value of x+y+z is _________.

Moderate

Solution

$\begin{array}{l} F_{21} = F_{21} \hat{e} = F_{21} \frac{A B}{| A B |} = F_{21} \frac{(\hat{i} + 2 \hat{j} + 2 \hat{k})}{\sqrt{1 + 4 + 4}} \\ = \frac{3}{3} (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k} \\ ∴ x = 1, y = 2, z = 2 \\ ∴ x + y + z = 5 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App