Question

Two metal spheres having charge densities \[5\mu C/{{m}^{2}}\] and $-2\mu C/{{m}^{2}}$ with radii 2mm and 1mm respectively are kept in a hypothetical closed surface. Calculate total normal electric induction over the closed surface.

Answer 1

Hint: The charge densities and radii of two metal spheres are given. First by using the formula for charge density, we can find the charge of two metal spheres. Later by substituting the value of charges in the equation to find total normal electric induction, we will get the solution.
Formula used:
Charge density, \[\sigma =\dfrac{q}{A}\]
Total normal electric induction, $\text{T}\text{.N}\text{.E}\text{.I}={{q}_{1}}+{{q}_{2}}$

Complete answer:
In the question it is said that there are two metal spheres and we are given the charge densities and radii of these spheres.
Let ‘$\sigma $’ be charge density and ‘${{\sigma }_{1}}$’ and ‘${{\sigma }_{2}}$’ be the charge densities of metal sphere 1 and 2 respectively.
 Thus we have,
${{\sigma }_{1}}=5\mu C/{{m}^{2}}$
${{\sigma }_{2}}=-2\mu C/{{m}^{2}}$
These are given in micro coulomb. Let us convert them into coulomb.
${{\sigma }_{1}}=5\times {{10}^{-6}}C/{{m}^{2}}$
${{\sigma }_{2}}=-2\times {{10}^{-6}}C/{{m}^{2}}$
Let ‘${{r}_{1}}$’ and ‘${{r}_{2}}$’ be the radii of the metal sphere 1 and 2 respectively. Then we are given,
${{r}_{1}}=2mm=2\times {{10}^{-3}}m$
${{r}_{2}}=1mm=1\times {{10}^{-3}}m$
We know that charge density is given by the equation,
\[\sigma =\dfrac{q}{A}\], where ‘q’ is the charge of the conductor and ‘A’ is the area.
From the above equation we can write the charge density of metal sphere 1 as,
${{\sigma }_{1}}=\dfrac{{{q}_{1}}}{{{A}_{1}}}$
From this equation we get charge of the first metal sphere as,
$\Rightarrow {{q}_{1}}={{A}_{1}}{{\sigma }_{1}}$
We know that the area of a sphere is $=4\pi {{r}^{2}}$.
Since we have metal spheres, the area of the metal sphere 1 will be,
${{A}_{1}}=4\pi {{r}_{1}}^{2}$
By substituting area in the equation we will get the charge of the metal sphere 1 as,
$\Rightarrow {{q}_{1}}=4\pi {{r}_{1}}^{2}\times {{\sigma }_{1}}$
Now we can substitute the values of radius and charge density of the first metal sphere in the above equation.
Thus we get,
$\Rightarrow {{q}_{1}}=4\pi {{\left( 2\times {{10}^{-3}} \right)}^{2}}\times 5\times {{10}^{-6}}$
By solving this, we get
$\Rightarrow {{q}_{1}}=2.513\times {{10}^{-10}}C$
Similarly we can write charge density of the metal sphere 2 as,
${{\sigma }_{2}}=\dfrac{{{q}_{2}}}{{{A}_{2}}}$
Area of the metal sphere 2 will be,
${{A}_{2}}=4\pi {{r}_{2}}^{2}$
Therefore we will get charge of the second metal sphere as,
$\Rightarrow {{q}_{2}}={{\sigma }_{2}}{{A}_{2}}$
$\Rightarrow {{q}_{2}}=4\pi {{r}_{2}}^{2}{{\sigma }_{2}}$
By substituting the values of charge density and radius of the metal sphere 2, we will get
$\Rightarrow {{q}_{2}}=4\pi {{\left( 1\times {{10}^{-3}} \right)}^{2}}\left( -2\times {{10}^{-6}} \right)$
By solving this,
$\Rightarrow {{q}_{2}}=-2.513\times {{10}^{-11}}C$
We need to find the total normal electric induction over the closed surface.
We know that total normal electric induction (T. N. E. I) over a closed surface is the sum of the total charges enclosed by that surface, i.e.
$\text{T}\text{.N}\text{.E}\text{.I}={{q}_{1}}+{{q}_{2}}$
By substituting for ${{q}_{1}}$ and ${{q}_{2}}$, we get
\[\text{T}\text{.N}\text{.E}\text{.I}=\left( 2.513\times {{10}^{-10}} \right)+\left( -2.513\times {{10}^{-11}} \right)\]
By calculating, we will get
$\Rightarrow \text{T}\text{.N}\text{.E}\text{.I=2}\text{.262}\times \text{1}{{\text{0}}^{-10}}N{{m}^{2}}/C$
Hence the total normal electric induction over a closed a closed surface is, $\text{2}\text{.262}\times \text{1}{{\text{0}}^{-10}}N{{m}^{2}}/C$

Note:
In the question it is said that the two metal spheres are kept in a hypothetical closed surface.
A hypothetical closed surface in other words is known as Gaussian surface.
We know that according to gauss’s law the net electric flux on a closed surface is the ratio of charge enclosed by that surface and permittivity. This closed surface where we apply gauss’s law is a Gaussian surface.