Three identical charges, each $2\mu \mathrm{C}$ lie at the vertices of a right angled triangle as shown in the figure. Forces on the charge at B due to the charges at A and C respectively are F₁ and F₂. The angle between their resultant force and F₂  is

Given data: according to the diagram, each $2\mu \mathrm{C}$ is located at the vertices of a right-angled triangle. Forces$F_1\text{ and }{F}_2$ are applied to the charge at B as a result of the charge at A and C , respectively.

Concept used:  Electric field and electric field lines.

Detailed solution: 

If the $F_1$ and $F_2$ makes an angle $\theta$ with force $F_2$.

 $\tan \left(\theta \right)=\frac{F_1}{F_2}$

 $F_1=k·\frac{q_1{q}_2}{3^2}$ 

 $q_1=q_2=q_3=2\mu C$ 

 $F_2=k\frac{q_1{q}_3}{4^2}$ 

 $\tan \left(\theta \right)=\frac{k{q}_1{q}_2/3^2}{k{q}_1{q}_3/4^2}=\frac{4^2}{3^2}=\frac{16}{9}$ 

 $\Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{16}{9}\right)$