Question:
On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by \(\frac{Ax}{(x^2+a^2)^{3/2}}\) 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 :
On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by \(\frac{Ax}{(x^2+a^2)^{3/2}}\) 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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Potential is related to the field by \(E=-\frac{dV}{dx}\), so integrate the field from \(x\) to infinity.
Updated On: Nov 5, 2026
- \(\frac{A}{(x^2+a^2)^{3/2}}\)
- \(\frac{A}{(x^2+a^2)^{1/2}}\)
- \(A(x^2+a^2)^{3/2}\)
- \(A(x^2+a^2)^{1/2}\)
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The Correct Option is B
Solution and Explanation
\(|V(x)|=\int_x^{\infty}E\,dx=A\int_x^{\infty}\frac{x\,dx}{(x^2+a^2)^{3/2}}\).
Put \(u=x^2+a^2\), \(x\,dx=du/2\): \(\int u^{-3/2}\frac{du}{2}=-u^{-1/2}\).
Evaluating from \(x\) to \(\infty\): \(|V|=A\left[-\frac{1}{\sqrt{x^2+a^2}}\right]_x^{\infty}=\frac{A}{(x^2+a^2)^{1/2}}\).
Answer: (2) \(\frac{A}{(x^2+a^2)^{1/2}}\)
Put \(u=x^2+a^2\), \(x\,dx=du/2\): \(\int u^{-3/2}\frac{du}{2}=-u^{-1/2}\).
Evaluating from \(x\) to \(\infty\): \(|V|=A\left[-\frac{1}{\sqrt{x^2+a^2}}\right]_x^{\infty}=\frac{A}{(x^2+a^2)^{1/2}}\).
Answer: (2) \(\frac{A}{(x^2+a^2)^{1/2}}\)
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