Question

$ABC$ is a right angled triangle in which $AB=3cm,BC=4cm$ 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, $F_{AB}=\frac{k{Q}_A{Q}_B}{r_{AB}^2}$
$\Rightarrow F_{AB}=\frac{k15\times 12\times 10^{-12}}{9\times 10^{-4}}=k20\times 10^{-8}N$
Similarly, $F_{BC}=\frac{k\times 20\times 12\times 10^{-12}}{16\times 10^{-4}}=k15\times 10^{-8}N$
Now, the resultant, $F_B=\sqrt{F_{AB}^2+F_{BC}^2}\left(\because \theta =90^{\circ }\right)$
$F_B=9\times 10\left[\sqrt{20^2+15^2}\right]=2250N$
Hence, the force acting on the charge at point $B$ is $2250N$