Question

Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Answer 1

Force acting on q due to Q₁ and Q₅ are opposite direction, so cancel to each other.

Force acting on q due to Q₃ is F₃ = `1/(4pi epsilon_0) ("qQ"_3)/"R"^2`

Force acting on q due to Q₂ and Q₄

Resolving in two-component method:

1. Vertical Component:
   Q₂ Sin θ and Q₄ Sinθ are equal and opposite direction, so they are cancel to each other.
2. Horizontal Component:
   Q₂ Sin θ and Q₄ cos θ are equal and same direction, so they can get added.
   F₂₄ = F_2q + F_4q = F₂ cos 55° + F₄ cos 45°
   F₂₄ = `1/(4pi epsilon_0) ("qQ"_3)/"R"^2 cos 45^circ + 1/(4pi epsilon_0) ("qQ"_4)/"R"^2 cos 45^circ`
   Resultant net force F
   
   F = F₃ + F₂₄ + F₁₅
   Q = Q₁ = Q₂ = Q₃ = Q₄ = Q₅
   cos 45° = `1/sqrt2`
   F = `"qQ"/(4pi epsilon_0 "R"^2) [1 + 1/sqrt2 + 1/sqrt2]`
   `= 1/(4pi epsilon_0) "qQ"/"R"^2 [1 + 2/sqrt2]`
   F = `1/(4pi epsilon_0) "qQ"/"R"^2 [1 + sqrt2]`N 
   Vector form:
   `vec"F" = 1/(4pi epsilon_0) "qQ"/"R"^2 (1 + sqrt2) "N"hat"i"`