Question

Two equally charged small balls placed at a fixed distance experiences a force F.A similar uncharged ball after touching one of them is placed at the middle point between the two balls.The force experienced by this ball is:

Answer 1

Answer:Let Q charged on two identical balls and separation between them is 2R

Then, repulsive force act between them , F = KQ²/(2R)² = KQ²/4R² ----(1)

Now, a 3rd ball is touched with one of them{ assume 1 St ball } ,

Then, charge will share equally to both the balls.

e.g., charge on 1st ball = Q/2

charge on 3rd ball = Q/2.

Now, 3rd ball is placed midpoint of 1st and 2nd ball .

so, Force act on 3rd ball due to 1st ball , F₁ = K(Q/2)(Q/2)/R² = KQ²/4R²

again, force act on 3rd ball due to 2nd ball , F₂ = K(Q/2)(Q)/R² = KQ²/2R²

Because both the forces are repulsive nature and F₂ > F₁

so, net force experienced by 3rd ball = F₂ - F₁ = KQ²/2R² - KQ²/4R² = KQ²/4R²

Now, from equation (1)

Net force experienced by 3rd ball = F { e.g., force experienced by 1st and 2nd ball }

Read more on Brainly.in - https://brainly.in/question/3198816#readmore

Explanation: