On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by ${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$ 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:
Solution
Given ${E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$<br><br>$\therefore$ ${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$<br><br>$\Rightarrow$ $$\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx} $$<br><br>$\Rightarrow$ $V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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