Question

Let $A B C$ is a right angled triangle in which $A B=3\, cm$ and $B C=4\, cm$ and $\angle A B C=90^{\circ}$. The three charges $+15,+12$ and $-20$ esu are placed on $A, B$ and $C$ respectively. The force acting on $B$ will be

• A. Zero
• B. 25 dyne
• C. 30 dyne
• D. 150 dyne

Answer 1

Answer: (B)

The given condition can be shown as

Net force on $B, F_{\text {net }}=\sqrt{F_{A}^{2}+F_{C}^{2}}$
$\therefore F_{A}=\frac{5 \times 12}{(3)^{2}}=20$ dyne
and $F_{C}=\frac{12 \times 20}{(4)^{2}}=15$ dyne
So, $F_{\text {net }}=\sqrt{(20)^{2}+(15)^{2}}=25$ dyne.