Question

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

• A. $\frac{Q^2}{\left(4\pi {\epsilon }_0{a}^2\right)}$
• B. $-Q^2\left(4\pi {\epsilon }_0{a}^2\right)$
• C. Zero
• D. $\frac{Q^2}{\left(2\pi {\epsilon }_0{a}^2\right)}$

Answer 1

Answer: (C)

Zero

Force due to both charge at B and C are equal in magnitude and equal to

$F_1=\frac{Q^2}{4\pi {ϵ}_o{a}^2}$

Now from figure the angle between two $F_1$ is $\theta ={120}^o$

Hence, net force $F=\sqrt{F_1^2+F_1^2+2F_1{F}_1\cos {120}^o}$

$⟹F=F_1$

$⟹F_1=\frac{Q^2}{4\pi {ϵ}_o{a}^2}$

Here the net force is parallel to BC so the force normal to BC is 0. As there is no component of force present in vertical direction