Question

A spherical conductor of radius $2cm$ is uniformly charged with $3nC$, what is the electric field at a distance of $3cm$ from the centre of the sphere?
$A)\text{ }3\times {{10}^{6}}V.{{m}^{-1}}$
$B)\text{ }3V.{{m}^{-1}}$
$C)\text{ }3\times {{10}^{4}}V.{{m}^{-1}}$
$D)\text{ }3\times {{10}^{-4}}V.{{m}^{-1}}$

Answer 1

Hint: A spherical conductor can be considered to be a point charge at the centre with the same magnitude of charge as that on the total conductor. Hence, we can find the total electric field using the formula for the electric field due to a point charge.
Formula used:
$\overrightarrow{E}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}\widehat{r}$

Complete step by step answer:
For the electric field outside the conductor, it can be considered to be equivalent to a point charge at the centre that has the same magnitude of charge as the whole conductor. Hence, we will get the required electric field by using the formula for that of a point charge.
The electric field $\overrightarrow{E}$ at a distance of $r$ from a point charge of charge magnitude $Q$ is given by
$\overrightarrow{E}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}\widehat{r}$ --(1)
where ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{m}^{-3}}k{{g}^{-1}}{{s}^{4}}{{A}^{2}}$ is the permittivity of free space and $\widehat{r}$ is the unit vector in the direction of the line joining the point charge to the required point.
Now, let us analyze the question.
As explained above, we will consider the conductor as a point charge at its centre with total charge equal to $Q=3nC=3\times {{10}^{-9}}C$ $\left( \because 1nC={{10}^{-9}}C \right)$
The distance of the point from the centre of the sphere is $r=3cm=3\times {{10}^{-2}}m$ $\left( 1cm={{10}^{-2}}m \right)$
Hence, using (1), we get the required electric field $\overrightarrow{E}$ at the point as
$\overrightarrow{E}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}\widehat{r}$
Hence, the magnitude $\overrightarrow{\left| E \right|}$ of the electric field will be
$\left| \overrightarrow{E} \right|=\left| \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}\widehat{r} \right|=\left| \dfrac{1}{4\pi {{\varepsilon }_{0}}} \right|\left| \dfrac{Q}{{{r}^{2}}} \right|\left| \widehat{r} \right|=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}$ $\left( \because \left| \widehat{r} \right|=1,\text{ since it is a unit vector} \right)$
Putting the values in the above question, we get,
$\left| \overrightarrow{E} \right|=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}=9\times {{10}^{9}}\times \dfrac{3\times {{10}^{-9}}}{{{\left( 3\times {{10}^{-2}} \right)}^{2}}}$ $\left( \because \dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}} \right)$
$\therefore \left| \overrightarrow{E} \right|=9\times {{10}^{9}}\times \dfrac{3\times {{10}^{-9}}}{\left( 9\times {{10}^{-4}} \right)}=3\times {{10}^{4}}V.{{m}^{-1}}$
Therefore, the required electric field at the point is $3\times {{10}^{4}}V.{{m}^{-1}}$.
Therefore, the correct option is $C)\text{ }3\times {{10}^{4}}V.{{m}^{-1}}$.

Note: Students must note that the result that the spherical charged conductor behaves as a point charge at the centre with the total charge of the conductor when the point in question, where the electric field has to be found is outside the conductor, comes from the application of Gauss’ Law. Similarly, for a point inside the conductor also, the electric field can be found out using Gauss’ Law. A point charge can also be considered at the centre in this case but it will have a magnitude of charge which will be a fraction of the total charge of the spherical conductor.