Two point charges 1C each separated by 1Km . Find the electrostatic force of
repulsion between them
Answers
Answered by
7
Answer:
- The electrostatic Repulsion (F) between the charges is 9 × 10³ N.
Given:
- Two charges have a magnitude of 1 C.
- Distance of separation (r) = 1 Km.
Explanation:
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 = 1 ]
⇒ F = (1 / 4 π ε₀) × (1 C × 1 C) / (1 km)²
⇒ F = (1 / 4 π ε₀) × (1 C × 1 C) / (10³ m)²
∵ [ 1 Km = 10³ m ] (S.I units)
⇒ F = (1 / 4 π ε₀) × 1 / 10⁶
We Know,
1 / 4 π ε₀ has an equivalent value of 9 × 10⁹
Substituting,
⇒ F = 9 × 10⁹ × 1 / 10⁶
⇒ F = 9 × 10⁹ × 1 × 10⁻⁶
⇒ F = 9 × 10⁹⁻⁶
⇒ F = 9 × 10³
⇒ F = 9 × 10³ N
∴ The electrostatic Repulsion (F) between the charges is 9 × 10³ N.
Answered by
4
- Electrostatic force between points is 9 × 10³ N
Given :
- First Charge (Q1) = 1 C
- Second Charge (Q2) = 1 C
- Distance between charges (d) = 1 km = 10³ m
To Find :
- Electrostatic Force between the charges
________________________
Solution :
Use Coulomb's Law
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