Solved Numerical Problem No.2 ( Unit-11: Electrostatics ); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

Charges 2 µC, -3 µC, and 4 µC are placed in the air at the vertices of an equilateral triangle of sides 10 cm. What is the magnitude of the resultant force acting on a 4 µC charge? (Answer 15.7 N) 

Given:

Let 

1st charge = q₁ = 2 µC = 2 ｘ 10⁻⁶ C 

2nd charge = q₂ = -3 µC = -3 ｘ 10⁻⁶ C 

3rd charge = q₃ = 4 µC = 4 ｘ 10⁻⁶ C 

Separation distance = r = 10 cm = 0.1 m


∴ Coulomb's Constant = K = `\frac {1}{4ㄫε_0}` = 9 ｘ 10⁹ Nm²C⁻²

To Find:

The magnitude of the resultant force acting on q₃  = F = ?

Solution:

The diagram can be drawn according to the given conditions are  as below

Where F₁ is the force of repulsions between charges q₁ and q₃ (as both are +ve charges) and F₂ is the force of attractions between charges q₂ and q₃ (as q₂ is -ve and q₃ is +ve charge)

F₁ cos 60° and F₂cos 60° are the horizontal X-components of the force F₁ and F₂ respectively. Similarly F₁ sin 60° and F₂ sin 60° are the vertical y-component of the force F₁ and F₂ respectively

Here we will first calculate F₁, the force of repulsion between charges q₁ and q₃ So using coulomb's Law formula


F₁ = `\frac {1}{4ㄫε_0}``\frac {q₁q₃}{r^2}`

by putting values

F₁= 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {2 ｘ 10⁻⁶ C ｘ 4 ｘ 10⁻⁶ C }{(0.1 m)^2}`


F₁= 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {8 ｘ 10⁻¹² C}{0.01 m^2}`


F₁ = `\frac {72 ｘ 10⁹⁻¹² Nm²}{0.01 m^2}`


F₁ = `\frac {72 ｘ 10⁻³ Nm²}{0.01 m^2}`


F₁ = 7200 ｘ 10⁻³ N


F₁ = 7.2 N ---------------(1).


and


F₂, the force of attraction between charges q₂ and q₃ So using coulomb's Law formula again


F₂ = `\frac {1}{4ㄫε_0}``\frac {q₂q₃}{r^2}`

by putting values

F₂ = 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {-3 ｘ 10⁻⁶ C ｘ 4 ｘ 10⁻⁶ C }{(0.1 m)^2}`


F₂ = 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {-12 ｘ 10⁻¹² C}{0.01 m^2}`


F₂ = - `\frac {108 ｘ 10⁹⁻¹² Nm²}{0.01 m^2}`


F₂ = -`\frac {108 ｘ 10⁻³ Nm²}{0.01 m^2}`


F₂ = -10800 ｘ 10⁻³ N


F₂ = -10.80 N ---------------(2).

The -ve sign just shows that it is an attraction force.

The resultant force F can be found by the formula 

F = `\sqrt {Fx^2 + Fy^2 }` ------------(3)


where  Fx = F₁x + F₂x  and Fy = F₁y + F₂y

Fx = F₁ cos 60° + F₂ cos 60°

Fx = 7.2 N ｘ 0.5  + (-10.80 N ) ｘ 0.5

Fx = 3.6 N + (-5.4 N )

Fx = -1.8 N ------------------(4)

and 

Fy = F₁ sin 60° - F₂ sin 60°  (opposite in direction)

Fy = 7.2 N ｘ 0.866  - (-10.80 N ) ｘ 0.866

Fy = 6.235 N + 9.353 N

Fy = 15.588 N --------------(5)

Putting Eqn (4)and (5) in eqn (3) we have

F = `\sqrt {Fx^2 + Fy^2 }`

F = `\sqrt {(-1.8 N)^2 + (15.588 N)^2 }`

F = `\sqrt {3.24N^2 + 242.986 N^2 }`

F = `\sqrt {246.226 N^2 }`

F = 15.693 N 

F = 15.7 N ------------------------Ans.

Thus the magnitude of the resultant force acting on q₃ is 15.7 N

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