ગાઉસનો ગુરૂત્વાકર્ષણનો નિયમ....

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Question

For electric field, we have: $|\overrightarrow{ E }|=\frac{ q }{4 \pi \epsilon_0 r ^2} \Rightarrow \oint \overrightarrow{ E } \cdot \overrightarrow{ S

Vedclass · Accepted Answer

$\oint {\mathop g\limits^ 	o \,.\,\mathop {ds}\limits^ 	o } \,\, = \, - 4\,G\pi m$

Vedclass · Answer

For electric field, we have: $|\overrightarrow{ E }|=\frac{ q }{4 \pi \epsilon_0 r ^2} \Rightarrow \oint \overrightarrow{ E } \cdot \overrightarrow{ S }=\frac{ q }{\epsilon_0}$. Similarly, for gravitational field, we have:

$|\vec{g}|=\frac{ Gm }{ r ^2} \Rightarrow G \equiv \frac{1}{4 \pi \epsilon_0} \Rightarrow 4 \pi G \equiv \frac{1}{\epsilon_0} \Rightarrow-\oint \overrightarrow{ g } \cdot \overrightarrow{ dS }=-4 \pi Gm 	ext {. }$

Please note that the gravitational field is always attractive in nature, hence we have the negative sign before the integral. Also note the similarity between gravitational field intensity and electric field intensity and between mass and charge.

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