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प्रश्न
A small hollow conducting sphere radius r1 is given a charge Q. It is surrounded by a concentric conducting spherical shell of inner radius r2 and outer radius r3, having charge −3q.
If a point charge 2q were kept at the centre, find:
- The electric flux through a concentric spherical Gaussian surfaces of radius x for (1) x < r1, and (2) r1 < x < r2.
- Electric field at a point distant x from the centre for (1) x > r3, and (2) r1 < x < r2.
- Surface charge density on inner surfaces of (1) sphere, and (2) shell.
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उत्तर
Given: Point charge at center = +2q
Inner sphere (r1) = Q
Outer sphere (r2, r3) = −3q
(I) Electric Flux (Gauss Law): The electric flux through a Gaussian surface is determined by the total enclosed charge q.
Φ = `q_"enclosed"/ε_0`
(1) For x < r1: The only charge enclosed is the point charge at the center (+2q).
Φ = `(2q)/ε_0`
(2) For r1 < x < r2: The enclosed charge is the point charge (2q) plus the total charge on the inner sphere (Q).
Φ = `(Q + 2q)/ε_0`
(II) Electric Field (E): The electric field at a distance x from the centre is given by
E =`1/(4 pi ε_0) (q_"enclosed")/x^2`
(1) For x > r3: The Gaussian surface encloses all charges: the point charge 2q, the inner sphere charge Q, and the outer shell charge −3q.
`q_"enclosed" = 2q + Q + (−3q)`
= Q − q
E = `1/(4 pi ε_0) (Q − q)/x^2`
= `(Q − q)/(4 pi ε_0 x^2)`
(2) For r1 < x < r2: The enclosed charge is the point charge 2q and the charge Q on the inner sphere.
`q_"enclosed" = 2q + Q`
= Q + 2q
E = `1/(4 pi ε_0) (Q + 2q)/x^2`
= `(Q + 2q)/(4 pi ε_0 x^2)`
(III) Surface Charge Density (σ)
σ = `"Induced Charge"/"Surface Area"`
(1) Inner surface of sphere (r1): To make the field zero within the conducting wall of the sphere, the charge on its inner surface must exactly neutralize the point charge 2q at the centre.
Induced charge is −2q.
`σ_(r_(1)) = (-2q)/(4pir_1^2)`
= `-(q)/(2pir_1^2)`
(2) Inner surface of shell (r2): To make the field zero within the conducting shell, the charge on its inner surface must neutralize the total charge inside it (the point charge 2q and the total charge Q on the inner sphere).
Induced charge is −(Q + 2q).
`σ_(r_(2)) = (-Q + 2q)/(4pir_2^2)`
