Two equal charges placed 3m apart repel each other with force 0.1gf. Calculate the
magnitude of either charge
Answers
Answer:
- The Magnitude of the charges (q) is 9.9 × 10⁻⁷ C
Given:
- Electrostatic force of attraction (F) = 0.1 g F
- Distance of separation (r) = 3 m.
- Let the two charges have a magnitude equal to q
Explanation:
First we need to convert Force into S.I units.
We Know,
⇒ 1 g F = 0.0098 N
⇒ 1 g F = 9.8 × 10⁻³ N
Now,
⇒ 0.1 g F = 0.1 × 9.8 × 10⁻³ N
⇒ 0.1 g F = 9.8 × 10⁻⁴ N
⇒ 0.1 g F = 9.8 × 10⁻⁴ N
∴ We got the value of force in S.I units.
From coulomb's law we know,
⇒ F = (1 / 4 π ε₀) × (q₁ q₂ / r²)
Where,
- F Denotes Electrostatic force of repulsion.
- ε₀ Denotes Permmittivity of free space.
- q₁ Denotes Charge.
- q₂ Denotes Charge.
- r Denotes Distance.
Now,
⇒ F = (1 / 4 π ε₀) × (q₁ q₂ / r²)
Substituting the values,
∵ [ q₁ = q₂ = q ]
⇒ 9.8 × 10⁻⁴ = (1 / 4 π ε₀) × (q × q) / (3 m)²
⇒ 9.8 × 10⁻⁴ = (1 / 4 π ε₀) × q² / 9 m²
We Know,
1 / 4 π ε₀ has an equivalent value of 9 × 10⁹
Substituting,
⇒ 9.8 × 10⁻⁴ = 9 × 10⁹ × q² / 9
⇒ 9.8 × 10⁻⁴ = 10⁹ × q²
⇒ q² = 9.8 × 10⁻⁴ / 10⁹
⇒ q² = 98 × 10⁻⁵ / 10⁹
⇒ q² = 98 × 10⁻⁵⁻⁹
⇒ q² = 98 × 10⁻¹⁴
⇒ q = √ 98 × 10⁻¹⁴
⇒ q = 10⁻⁷ × √ 98
⇒ q = 10⁻⁷ × 9.89
⇒ q = 9.89 × 10⁻⁷
⇒ q ≈ 9.9 × 10⁻⁷ C
∴ The Magnitude of the charges (q) is 9.9 × 10⁻⁷ C.
- Force of attraction (F) = 0.1 g F
- Distance of separation (r) = 3 m.
___________
Assuming that the two charges each have a magnitude of Q;
___________
Converting the values to SI Units ;
⇒ 0.1 g F = 9.8 × 10⁻⁴ N
- F Denotes Electrostatic force of repulsion.
- q₁ Denotes Charge.
- q₂ Denotes Charge.
r Denotes Distance.
Substituting the values, and solving we get :
The Magnitude of the charges (q) is 9.9 × 10⁻⁷ C.