Question

8. Two point charges Q1 and Q2 are 3m away from each other. The some two charges is 20uc they
repel each other with a force of 0.075N. Find the two charges.​

Answer 1

Answer:

Q₁ = 15 × 10⁻⁶ C = 15 μC

Q₂ = 5 × 10⁻⁶ C = 5 μC

Explanation:

Given----> Two charges Q₁ and Q₂ are 3m away from each other and sum of two charges is 20μC and they repel each other with a force of 0.075 N

To find ----> Value of two charges

Solution----> 1) Plzz refer the attachement

2) ATQ, Q₁ + Q₂ = 20μC

=> Q₁ + Q₂ = 20 × 10⁻⁶ C

=> Q₁ + Q₂ = 20 × 10⁻⁶ ................( 1 )

3) Then we apply formula of electrostatic force

F = 1 / 4πε₀ Q₁ Q₂ / r²

Putting F = 0.075 N , r = 3 m

by applying this we get ,

Q₁ Q₂ = 75 × 10⁻¹²

4) We know that,

( Q₁ - Q₂ )² = ( Q₁ + Q₂ )² - 4 Q₁ Q₂

Applying it we get,

Q₁ - Q₂ = 10 × 10⁻⁶ ...................( 2 )

5) Now we have two equations ,

Q₁ + Q₂ = 20 × 10⁻⁶ ...............( 1 )

Q₁ - Q₂ = 10 × 10⁻⁶ ................( 2 )

Adding ( 1 ) and ( 2 ) , we get,

Q₁ + Q₂ + Q₁ - Q₂ = 20 × 10⁻⁶ + 10 × 10⁻⁶

=> 2 Q₁ = 30 × 10⁻⁶

=> Q₁ = 15 × 10⁻⁶ C

Putting Q₁ = 15 × 10⁻⁶ in equation ( 1 ) , we get,

=> Q₁ + Q₂ = 20 × 10⁻⁶

=> Q₂ = 20 × 10⁻⁶ - Q₁

=> Q₂ = 20 × 10⁻⁶ - 15 × 10⁻⁶

=> Q₂ = 5 × 10⁻⁶ C