A spherical conductor of radius $12 \;cm$ has a charge of $1.6 \times 10^{-7} \;C$ distributed uniformly on its surface. What is the electric field
$(a)$ inside the sphere
$(b)$ just outside the sphere
$(c)$ at a point $18\; cm$ from the centre of the sphere?
A spherical conductor of radius $12 \;cm$ has a charge of $1.6 \times 10^{-7} \;C$ distributed uniformly on its surface. What is the electric field
$(a)$ inside the sphere
$(b)$ just outside the sphere
$(c)$ at a point $18\; cm$ from the centre of the sphere?
$(a)$ Radius of the spherical conductor, $r=12 \,cm =0.12\, m$
Charge is uniformly distributed over the conductor, $q=1.6 \times 10^{-7}\, C$
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. $E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$
Where, $\varepsilon_{0}=$ Permittivity of free space and $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}\, Nm ^{2} \,C ^{-2}$
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 $10^{5} \,N\, C^{-1}$
$(c)$ Electric field at a point $18\, m$ from the centre of the sphere $= E _{1}$ Distance of the point from the centre, $d =18 \,cm =0.18\, m$
$E_{1}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}$$=4.4 \times 10^{4} \,N\,C ^{-1}$
Therefore, the electric field at a point $18\, cm$ from the centre of the sphere is $4.4 \times 10^{4} \,N\, C^{-1}$
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