Three charges are present on the three vertices of a triangle with charges -3q,-4q and +2q respectively. Find the total force exerted by the charges and the direction of the force
Answers
Answer:
hree charges 10ue , -10uc and 10uc are placed at the vertices of an equilateral triangle of side 10 m . Find the electrostatics potential energy.What is the work done to dissociate the system of these three charges.
Explanation:
Given:
Three charges are present on the three vertices of a triangle with charges -3q,-4q and +2q respectively.
To find:
Find the total force exerted by the charges and the direction of the force
Solution:
From given, we have,
The triangle is an equilateral triangle.
we use the formula,
F = 1/4 π∈₀ (q₁q₂)/d²
The force between charges -3q and -4q is given by,
F₁ = 1/4 π∈₀ (3q×4q)/d²
F₁ = 1/ π∈₀ (3q²)/d²
3F₁ = 1/ π∈₀ (q²/d²)
The force between charges -4q and +2q is given by,
F₂ = 1/4 π∈₀ (2q×4q)/d²
F₂ = 1/ π∈₀ (2q²)/d²
2F₂ = 1/ π∈₀ (q²/d²)
∴ 3F₁ = 2F₂
, so the angle between forces F₁ and F₂ is 120°
F = √[F₁² + F₂² + 2F₁F₂cos 120°]
F = √[F₁² + (3/2F₁)² + 2F₁(3/2F₁)cos 120°]
F = √[F₁² + 9/4F₁² + 3F₁²cos 120°]
F = √[F₁² + 9/4F₁² - 3/2F₁²]
F = √[7/4F₁²]
F = √7/2(F₁)
F = √7/2 [1/ π∈₀ (3q²)/d²]
∴ F = 3√7 [1/ 2π∈₀ (q²/d²)]