Question

two non conducting solid sphered having charge densitied P1 and P2 and raddi R and 2R are touching each other. net eelctric field at a ditance 2r from centre of smaller sphere on line joining the centres of the 2 sphered is 0, then P1/P2=?

Answer 1

4

Answer 2

Solution:

Let the centers of the two spheres be A (radius R, density P₁) and B (radius 2R, density P₂) with AB = R + 2R = 3R. Choose point P on line AB such that AP = 2R; then BP = 3R – 2R = R.

1. Field due to sphere A (outside):

   Total charge Q₁ = (4/3)πR³P₁.

   At P (distance 2R from A), by Coulomb’s law,

   E₁ = (1/(4πε₀))·(Q₁/(2R)²) = (1/(4πε₀))·((4/3)πR³P₁/(4R²))

   = (R P₁)/(12ε₀).

   Direction: from A to P.

2. Field due to sphere B (inside):

   For a uniformly charged sphere, inside the field is

   E = (P₂·r)/(3ε₀), where r is the distance from B. At P, r = R.

   Thus, E₂ = (P₂ R)/(3ε₀).

   Since P is inside sphere B, the field at P points radially outward from B.

   But here, since B lies beyond P, the direction of E₂ is from B to P (i.e. opposite to the direction from A to B).

3. Net Field at P:

   Assign positive direction from A to B. Then,

   E_net = E₁ (right) – E₂ (left) = (R P₁)/(12ε₀) – (R P₂)/(3ε₀).

   Setting E_net = 0 gives:

   (R P₁)/(12ε₀) = (R P₂)/(3ε₀) → P₁/12 = P₂/3 → P₁ = 4P₂.

Thus, P₁/P₂ = 4.

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Core Explanation:

• Calculate Q₁ = (4/3)πR³P₁.
• At a point 2R from A, electric field E₁ = (R P₁)/(12ε₀).
• For sphere B, point P lies inside; hence, E₂ = (P₂·R)/(3ε₀) with opposite direction.
• Equate E₁ and E₂ to get P₁ = 4P₂.