Question

Charges $25Q,9Q$ and $Q$ are placed at point $ABC$ such that $AB=4m$, $BC=3m$ and angle between $AB$ and $BC$ is ${90}^{\circ }$. then force on the charge $C$ is:

• A. Zero
• B. $\frac{q^2}{\pi {\in }_0\sqrt{5}}$
• C. $\frac{2Q^2}{\pi {\in }_0}$
• D. $\frac{5Q^2}{4\pi {\in }_0}$

Answer 1

The correct option is B $\frac{q^2}{\pi {\in }_0\sqrt{5}}$

R.E.F image

Let us define,

$q_1=25;q_2=9q;q_3=q$ placed at A, B and C respectively, as shown in figure.

The total force on $q_3$ will be the vector sum of force on $q_3$ due to $q_2$ (ie $F_{32}$ ) and force on $q_3$ due to $q_1$ (ie $F_{31}$).

$F_{32}=\frac{q_3{q}_2}{4\pi {ϵ}_0(BC{)}^2}=\frac{9q^2}{4\pi {ϵ}_0\times 9}=\frac{q^2}{4\pi {ϵ}_0}$

$F_{31}=\frac{q_3{q}_1}{4\pi {ϵ}_0\left(\right(AB{)}^2+\left(BC{)}^2\right)}=\frac{25q^2}{4\pi {ϵ}_0\times 25}=\frac{q^2}{4\pi {ϵ}_0}$

$F_{total}=\sqrt{F_{32}^2+F_{31}^2+2F_{32}{F}_{31}cos\theta }$

here, $cos\theta =\frac{3}{5}$

$\Rightarrow F_{total}=\frac{q^2}{4\pi {ϵ}_0}\sqrt{1+1+2\times \frac{3}{5}}$

$\Rightarrow F_{total}=\frac{q^2}{\sqrt{5}\pi {ϵ}_0}$