Question

$ABC$ is a right angled triangle in which $AB=3 \,cm , BC=4 \,cm$ and right angle is at $B$ Three charges $+15 \mu C ,+12 \mu C$ and $-20 \,\mu C$ are placed respectively at $A, B$ and $C $ The force acting on the charge at $B$ is

• A. 1250 N
• B. 3500 N
• C. 1200 N
• D. 2250 N

Answer 1

Answer: (D)

According to the question, $3 $ charge particles are placed at the vertices of a right angle triangle $ABC$ as shown in the figure below,

Where, $Q_{A}=+15 \,\mu C $,
$Q_{B} =12\, \mu C $
and $Q_{C}=-20 \,\mu C$
Now, the force, $F_{A B}=\frac{k Q_{A} Q_{B}}{r_{A B}^{2}}$
$\Rightarrow \, F_{A B}=\frac{k 15 \times 12 \times 10^{-12}}{9 \times 10^{-4}}=k 20 \times 10^{-8} \,N$
Similarly, $F_{B C}=\frac{k \times 20 \times 12 \times 10^{-12}}{16 \times 10^{-4}}=k\, 15 \times 10^{-8} \,N$
Now, the resultant, $F_{B}=\sqrt{F_{A B}^{2}+F_{B C}^{2}} \,\left(\because \,\theta=90^{\circ}\right)$
$F_{B}=9 \times 10\left[\sqrt{20^{2}+15^{2}}\right]=2250 \,N$
Hence, the force acting on the charge at point $B$ is $2250\,N$