Question

3 point charges are kept at the vertices of an equilateral triangle of side 20cm. what should be the magnitude of the charge to be placed at the mid point of side BC so that the charge A remains in equilibrium ​

Answer 1

Given:

3 point charges are kept at the vertices of an equilateral triangle of side 20cm.  

To find:

What should be the magnitude of the charge to be placed at the mid point of side BC so that the charge A remains in equilibrium ​

Solution:

We use the formula,

F = 1/4π∈₀ q₁q₂/d²

From given, we have,

F_{AB} = 1/4π∈₀ q₁q₂/AB²

F_{AB} = 9 × 10⁹ [(2×10^{-6}) (3×10^{-6})]/(0.20)²

F_{AB} = 1.35  N along AB

F_{AC} = 1/4π∈₀ q₁q₂/AC²

F_{AC} = 9 × 10⁹ [(2×10^{-6}) (3×10^{-6})]/(0.20)²

F_{AC} = 1.35  N along AB

F_{AO} = 1/4π∈₀ q₁q₂/AB²

F_{AO} = 9 × 10⁹ [(2×10^{-6}) (q)]/(√3×10^{-1})²

F_{AO} = 6 × 10⁵ q  along OA

F_{AB} cos 30° + F_{AC} cos 30° = F_{AO}

1.35 × √3/2 + 1.35 × √3/2 = 6 × 10⁵ q

∴ q =   1.35√3/6 × 10⁵ = 0.3897 × 10^{-5} C

Therefore, magnitude of the charge to be placed at the mid point of side BC so that the charge A remains in equilibrium ​is 3.897 μC