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  • On the x-axis and at a distance x from the origin, the gravitational field due to mass distribution is given by <img alt="\frac{Ax}{(x^{2}+a^{2})^{\frac{3}{2}}}" src="http://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7BAx%7D%7B%28x%5E%7B2%7D&am

On the x-axis and at a distance x from the origin, the gravitational field due to mass distribution is given by \frac{Ax}{(x^{2}+a^{2})^{\frac{3}{2}}} in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:
Option: 1 \frac{A}{(x^{2}+a^{2})^{\frac{1}{2}}}
Option: 2 \frac{A}{(x^{2}+a^{2})^{\frac{3}{2}}}
Option: 3 A(x^{2}+a^{2})^{\frac{1}{2}}
Option: 4 A(x^{2}+a^{2})^{\frac{3}{2}}

Answers (1)

best_answer

\begin{array}{l} \text { Given } \\ \mathrm{E}_{\mathrm{G}}=\frac{\mathrm{Ax}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{3 / 2}}, \mathrm{~V}_{\infty}=0 \\ Using \\ \int_{\mathrm{V}_{\infty}}^{\mathrm{V}_{x}} \mathrm{dV}=-\int_{\infty}^{\mathrm{x}} \overrightarrow{\mathrm{E}}_{\mathrm{G}} \cdot \overrightarrow{\mathrm{d}}_{\mathrm{x}} \\ \\ \mathrm{V}_{\mathrm{x}}-\mathrm{V}_{\infty}=-\int_{\infty}^{\mathrm{x}} \frac{\mathrm{Ax}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{3 / 2}} \mathrm{dx} \\ \end{array}

\\ \text { put } \mathrm{x}^{2}+\mathrm{a}^{2}=\mathrm{z} \\ 2 \mathrm{x} \mathrm{dx}=\mathrm{d} \mathrm{z}

So

\begin{array}{c} \mathrm{V}_{\mathrm{x}}-0=-\int_{\infty}^{\mathrm{x}} \frac{\mathrm{Adz}}{2(\mathrm{z})^{3 / 2}}=\left[\frac{\mathrm{A}}{\mathrm{z}^{1 / 2}}\right]_{\infty}^{\mathrm{x}}=\left[\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}}\right]_{\infty}^{\mathrm{x}} \\ \\ \mathrm{V}_{\mathrm{x}}=\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}}-0=\frac{\mathrm{A}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{1 / 2}} \end{array}

 

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avinash.dongre

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