Question

Three charges are placed at the vertices of an equilateral triangle of side ‘a’ as shown in the following figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is

• A. $\frac{Q^2}{4\pi {ϵ}_0{a}^2}$
  
• B. $\frac{-Q^2}{4\pi {ϵ}_0{a}^2}$
  
• C. $Zero$
  
• D. $\frac{Q^2}{2\pi {ϵ}_0{a}^2}$

Answer 1

Answer: (C)

$Zero$

According to the Coulomb's law

We know,

$|\overrightarrow{F_B}|=|\overrightarrow{F_C}|=k.\frac{Q^2}{a^2}$

Force($F_C$ ) between the charges placed at A and C is in the direction away from the charge and force($F_B$ ) between the charges placed at the A and B is in the direction towards the B as shown in the figure.

After resolving both the forces in X and Y direction forces in the X direction will add up and forces in the Y direction will cancel each other as shown in the figure.

Hence force experienced by the charge at A in the direction normal to BC is zero.