Ncert Exercises & Solutions

A spherical conductor of radius \(12~\text{cm}\) carries a uniformly distributed charge of \(1.6 \times 10^{-7}~\text C\) on its surface. What are the electric field strengths \(E_{in}\)(inside the sphere), \(E_{out}\) (just outside the sphere), and \(E_{at~18~\text{cm}}\) (at a distance of 18 cm from the center of the sphere), respectively?

1.	\(E_{in} = 0~\text{N/C};~E_{out} =8.5 \times 10^4~\text{N/C};~E_{at~18~\text{cm}} = 3.8\times 10^4~\text{N/C}\)
2.	\(E_{in} = 2.0\times 10^4~\text{N/C};~E_{out} = 0~\text{N/C};~E_{at~18~\text{cm}} = 3.8\times 10^4~\text{N/C}\)
3.	\(E_{in} = 0~\text{N/C};~E_{out} =1.0 \times 10^5~\text{N/C};~E_{at~18~\text{cm}} = 4.4\times 10^4~\text{N/C}\)
4.	\(E_{in} = 0~\text{N/C};~E_{out} = 9.5 \times10^4~\text{N/C};~E_{at~18~\text{cm}} = 4.2\times 10^4~\text{N/C}\)

(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 x

10^{- 7}

C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.

(b) Electric field E just outside the conductor is given by the relation,

E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}

Where, E = Permittivity of free space and

\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}

Therefore,

E = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 7}}{(0.12)^{2}} = 10^{5} N C^{- 1}

Therefore, the electric field just outside the sphere is

10^{5} {NC}^{- 1}

(c)Electric field at a point 18 m from the centre of the sphere E

The distance of the point from the centre, d = 18 cm 0.18 m

E_{1} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{d^{2}} = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 7}}{(1.8 \times 10^{- 2})^{2}} = 4.4 \times 10^{4} N C^{- 1}

Therefore, the electric field at a point 18 cm from the center of the sphere is 4.4 x

10^{4} {NC}^{- 1}

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