On the x− axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax(x2+a2)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 :

Given : Gravitational field,EG=Ax(x2+a2)3/2,V∞=0∫V∞VxdV=−∫∞xE→G.d→x ⇒Vx−V∞=−∫∞xAx(x2+a2)3/2dx∴Vx=A(x2+a2)1/2−0=A(x2+a2)1/2

On the x− axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax(x2+a2)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 :

Given : Gravitational field,EG=Ax(x2+a2)3/2,V∞=0∫V∞VxdV=−∫∞xE→G.d→x ⇒Vx−V∞=−∫∞xAx(x2+a2)3/2dx∴Vx=A(x2+a2)1/2−0=A(x2+a2)1/2

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On the $x -$ axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ 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 :

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$\frac{A}{{(x^{2} + a^{2})}^{1 / 2}}$

$\frac{A}{{(x^{2} + a^{2})}^{3 / 2}}$

$A {(x^{2} + a^{2})}^{1 / 2}$

$A {(x^{2} + a^{2})}^{3 / 2}$

answer is A.

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Detailed Solution

Given : Gravitational field, $E_{G} = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}, V_{\infty} = 0$

$\int_{V_{\infty}}^{V_{x}} d V = - \int_{\infty}^{x} {\vec{E}}_{G} . {\vec{d}}_{x} \Rightarrow V_{x} - V_{\infty} = - \int_{\infty}^{x} \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}} d x$

$∴ V_{x} = \frac{A}{{(x^{2} + a^{2})}^{1 / 2}} - 0 = \frac{A}{{(x^{2} + a^{2})}^{1 / 2}}$

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On the x− axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax(x2+a2)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 :

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