Answer
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Hint: According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$
Where the electric field, $E$ is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$
By comparing the equation for electric field and gravitational field we can arrive at the gauss law for gravitational field.
Complete step by step answer:
According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$ (1)
Where the electric field is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$ (2)
Similarly, we can write the gauss law for gravitational field.
We know gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$ (3)
Now compare equation (3) and (2)
We can see that $q$in electric field is analogous to$m$ in gravitational field.
Also, the constants $\dfrac{1}{{4\pi {\varepsilon _0}}}$ in electric field is analogous to the gravitational constant $G$in gravitational field. Now we can equate these constants.
$
\dfrac{1}{{4\pi {\varepsilon _0}}} = G \\
\dfrac{1}{{{\varepsilon _0}}} = 4\pi G \\
$
So, let us replace the gravitational analogues in the equation (1)
$
\oint {E.ds} = \dfrac{1}{{{\varepsilon _0}}} \times q \\
\oint {g.ds} = 4\pi G \times m \\
$
The gravitational force is always an attractive force. Hence, we have to consider a negative sign in the equation of gauss law for the gravitational field.
Thus, our final answer is option B, $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = 4\pi mG$
Note: It is important to note that Gravitational force is always attractive in nature,but $4\pi mG$ is always a positive quantity and hence we include a negative sign in the left hand side of the equation to maintain the sign convention(negative field for attraction).
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$
Where the electric field, $E$ is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$
By comparing the equation for electric field and gravitational field we can arrive at the gauss law for gravitational field.
Complete step by step answer:
According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$ (1)
Where the electric field is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$ (2)
Similarly, we can write the gauss law for gravitational field.
We know gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$ (3)
Now compare equation (3) and (2)
We can see that $q$in electric field is analogous to$m$ in gravitational field.
Also, the constants $\dfrac{1}{{4\pi {\varepsilon _0}}}$ in electric field is analogous to the gravitational constant $G$in gravitational field. Now we can equate these constants.
$
\dfrac{1}{{4\pi {\varepsilon _0}}} = G \\
\dfrac{1}{{{\varepsilon _0}}} = 4\pi G \\
$
So, let us replace the gravitational analogues in the equation (1)
$
\oint {E.ds} = \dfrac{1}{{{\varepsilon _0}}} \times q \\
\oint {g.ds} = 4\pi G \times m \\
$
The gravitational force is always an attractive force. Hence, we have to consider a negative sign in the equation of gauss law for the gravitational field.
Thus, our final answer is option B, $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = 4\pi mG$
Note: It is important to note that Gravitational force is always attractive in nature,but $4\pi mG$ is always a positive quantity and hence we include a negative sign in the left hand side of the equation to maintain the sign convention(negative field for attraction).
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