Question

two identical balls having like charges and placed at a certain distance apart repel each other with a certain foods they are brought in contact and then moved a path to a distance equal to half the initial separation for stop the force of repulsion between them increases 4.5 times in comparison with initial value the ratio of the initial charge of the wall is

Answer 1

Answer:

Explanation:

Let there are two balls A and B

Charge on ball A = Q₁

Charge on ball B = Q₂

Distance between ball A and ball B = r

Then, force act between them , F = KQ₁Q₂/r² -------(1)

Now, balls are brought in contact . Then moved to a apart to a distance equal to half of intial.

You know, after contacting balls have equal amount of charge.

And magnitude of charge on each = (Q₁ + Q₂)/2

distance between them = r/2

Now, force between them , F' = K(Q₁ + Q₂)²/4 × r²/4 = K(Q₁ + Q₂)²/r² -----(2)

A/C to question ,

F = 4.5F'

⇒9KQ₁Q₂/r² = 2K(Q₁ + Q₂)²/r²

⇒9Q₁.Q₂ = 2(Q₁ + Q₂)²

⇒9Q₁Q₂ = 2Q₁² + 2Q₂² + 4Q₁Q₂

⇒ 5Q₁Q₂ = 2Q₁² + 2Q₂²

⇒ 5 = 2k + 2/k [ let k = Q₁/Q₂ ]

⇒2k² - 5k + 2 = 0

⇒ (k -2)(2k - 2) = 0

K = 2 or 1/2

Hence , Q₁/Q₂ = 2 or 1/2