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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"/(4piin_0"r"^2)`
Where,
`in_0` = Permittivity of free space
`1/(4piin_0) = 9 xx 10^9 "N m"^2 "C" ^-2`
∴ `"E" = (1.6 xx 10^-7 xx 9 xx 10^-9)/(0.12)^2`
= `10^5 "N C"^-1`
Therefore, the electric field just outside the sphere is `10^5 "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"/(4piin_0"d"^2)`
= `(9 xx 10^9 xx 1.6 xx 10^-7)/(18 xx 10^-2)^2`
= `4.4 xx 10^4 "N"/"C"`
Therefore, the electric field at a point 18 cm from the centre of the sphere is `4.4 xx 10^4 "N"/"C"`.
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संबंधित प्रश्न
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