Ncert Exercises & Solutions

Q 2.4) A spherical conductor of radius 12 cm has a charge of 1.6 x 10^-7C distributed uniformly on its surface. What is the electric field

(1) inside the sphere.

(2) just outside the sphere.

(3) at a point 18 cm from the centre of sphere.

(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}

$10^{- 7}$ 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}}

$E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$
Where, E = Permittivity of free space and

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

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

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

$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 7}}{(0.12)^{2}} = 10^{5} N C^{- 1}$
Therefore, the electric field just outside the sphere is

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

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

(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}

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

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

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

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

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