Question

Three equal charges, each having a magnitude of $2.0\times {10}^{-6}C$ are placed at the three corners of a right angled triangle of sides $3cm,$ $4cm$ and $5cm.$ The force (in magnitude) on the charge at the right angled corner is :

• A. $50N$
• B. $26N$
• C. $29N$
• D. $45.9N$

Answer 1

The correct option is D $45.9N$

Charge at each corner $q=2.0\times {10}^{-6}C$

Force between two charges $F=\frac{k{q}^2}{r^2}$

where $r$ is the distance between them and $k=9\times {10}^9$ units.

Force on the charge at the right angled corner $|F|=\sqrt{F_x^2+F_y^2}$

$F_x=\frac{k{q}^2}{(0.04{)}^2}$ $\hat{x}$

$\therefore$ $F_x=\frac{(9\times {10}^9)\times (2.0\times {10}^{-6}{)}^2}{(0.04{)}^2}=22.5N$ $\hat{x}$

$F_y=\frac{k{q}^2}{(0.03{)}^2}$ $\hat{y}$

$\therefore$ $F_y=\frac{(9\times {10}^9)\times (2.0\times {10}^{-6}{)}^2}{(0.03{)}^2}=40N$ $\hat{y}$

Thus $|F|=\sqrt{F_x^2+F_y^2}$

Or $|F|=\sqrt{(22.5{)}^2+(40{)}^2}$

$⟹$ $|F|=45.9N$