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 $4.4 \times 10^{4} \,N\, C^{-1}$
Similar Questions
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho$ (r) [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure The electric field is only along rhe radial direction.
Figure:$Image$
$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is
$(A)$ independent of a
$(B)$ directly proportional to a
$(C)$ directly proportional to $\mathrm{a}^2$
$(D)$ inversely proportional to a
$2.$ For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.
$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$
Give the answer question $1,2$ and $3.$
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho$ (r) [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure The electric field is only along rhe radial direction.
Figure:$Image$
$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is
$(A)$ independent of a
$(B)$ directly proportional to a
$(C)$ directly proportional to $\mathrm{a}^2$
$(D)$ inversely proportional to a
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$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.
$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$
Give the answer question $1,2$ and $3.$
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