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प्रश्न
A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7 C distributed uniformly on its surface. What is the electric field
- inside the sphere
- just outside the sphere
- at a point 18 cm from the centre of the sphere?
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उत्तर
(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10−7 C
The electric field inside a spherical conductor is zero. This is because if a field is inside the conductor, then charges will move to neutralize it.
(b) Electric field E just outside the conductor is given by the relation,
`E = q/(4piε_0r^2)`
Where,
`ε_0` = Permittivity of free space
`1/(4piε_0) = 9 xx 10^9` N m2 C−2
∴ `E = (1.6 xx 10^-7 xx 9 xx 10^9)/(0.12)^2`
= `(14.4 xx 10^2)/0.0144`
= 1000 × 102
= 105 N C−1
Therefore, the electric field just outside the sphere is 105 N C−1.
(c) Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
`E_1 = q/(4piε_0d^2)`
= `(9 xx 10^9 xx 1.6 xx 10^-7)/(18 xx 10^-2)^2`
= `(14.4 xx 10^2)/(324 xx 10^-4)`
= 0.044 × 106
= 4.4 × 104 N C−1
Therefore, the electric field at a point 18 cm from the centre of the sphere is 4.4 × 104 N C−1.
संबंधित प्रश्न
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
`(vec"E"_2 - vec"E"_1).hat"n" = sigma/in_0`
Where `hat"n"` is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of `hat"n"` is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ `hat"n"/in_0`
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
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