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

Question

$A B C$ is a right angled triangle in which $A B=3 \mathrm{~cm}, B C=4 \mathrm{~cm}$ and right angle is at $B$. Three charges $+15 \mu \mathrm{C},+12 \mu \mathrm{C}$ and $-20 \mu \mathrm{C}$ are placed respectively at $A, B$ and $C$. The force acting on the charge at $B$ is, $1250 \mathrm{~N}$, $3500 \mathrm{~N}$, $1200 \mathrm{~N}$, $2250 \mathrm{~N}$

MARKS App · Accepted Answer

According to the question, 3 charge particles are placed at the vertices of a right angle triangle $A B C$ as shown in the figure below, Where, $\begin{aligned} & Q_A=+15 \mu \mathrm{C}, \ & Q_B=12 \mu \mathrm{C} \ & Q_C=-20 \mu \mathrm{C} \end{aligned}$ and $Q_C=-20 \mu \mathrm{C}$ Now, the force, $F_{A B}=\frac{k Q_A Q_B}{r_{A B}^2}$ $\Rightarrow \quad F_{A B}=\frac{k 15 \times 12 \times 10^{-12}}{9 \times 10^{-4}}=k 20 \times 10^{-8} \mathrm{~N}$ Similarly, $F_{B C}=\frac{k \times 20 \times 12 \times 10^{-12}}{16 \times 10^{-4}}=k 15 \times 10^{-8} \mathrm{~N}$ Now, the resultant, $F_B=\sqrt{F_{A B}^2+F_{B C}^2} \quad\left(\because \theta=90^{\circ}\right)$ $F_B=9 \times 10\left[\sqrt{20^2+15^2}\right]=2250 \mathrm{~N}$ Hence, the force acting on the charge at point $B$ is $2250 \mathrm{~N}$. So, the correct option is $(d)$.

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