Animated Solution for Physics - Gravitation: 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

Imagine you are walking along the x-axis, and you can feel the gravitational pull from some mysterious mass distribution. The strength of this pull, the gravitational field, is given by the mathematical function $E_g=\frac{Ax}{(x^2+a^2{)}^{3/2}}$. If we visualize this, the field starts at zero at the origin, reaches a maximum, and then slowly fades away as we move towards infinity. Our mission is to find the gravitational potential $V(x)$ at any point $x$.

To build a bridge between the field we feel and the potential energy stored, we need our master tool from calculus. The fundamental relationship states that the small change in potential, $dV$, is equal to the negative dot product of the gravitational field and the small displacement $dx$:

Why the negative sign? Because potential naturally decreases as you move in the direction of the gravitational field! To find the total potential difference, we must integrate this expression from our reference point, which is infinity, all the way down to our point $x$.

The problem gives us a beautiful boundary condition: the potential at infinity is exactly zero ($V_{\infty }=0$). This makes our life so much easier! We substitute zero for $V$ at infinity and plug in the expression for our gravitational field. We are now looking at a definite integral from infinity to $x$:

Geometrically, this integral represents the area under our field curve from $x$ to infinity.

Don't get intimidated by this integral. Look closely at the numerator. We have an $xdx$ sitting right there. This is a massive hint! Whenever you see a function and its derivative present in the same integral, it's a perfect candidate for the substitution method. Let's introduce a new variable, $t$, and set it equal to the expression inside the bracket:

Differentiating both sides, we get $dt=2xdx$. Rearranging this, we find that $xdx=\frac{dt}{2}$.

Silly mistake alert! When you change the variable in a definite integral, you absolutely must change the limits of integration as well. When our lower limit $x$ approaches $\infty$, our new variable $t$ also approaches $\infty$. When $x$ is simply $x$, $t$ becomes $x^2+a^2$. Now our boundaries are perfectly translated into the world of $t$.

Let's rewrite our entire integral in terms of $t$. We replace the $xdx$ in the numerator with $\frac{dt}{2}$. The denominator simply becomes $t^{3/2}$. We can pull the constant $A$ and the factor of $\frac{1}{2}$ completely outside the integral:

According to the power rule for integration, we add one to the exponent, making it $-1/2$, and then we divide by that exact same new exponent. Dividing by $-1/2$ is the same as multiplying by $-2$. The negative sign from our formula and the negative sign from the integration cancel each other out. The twos also cancel perfectly!

We are at the finish line. Let's bring back our original variables and evaluate the limits. We plug in our upper limit, $x^2+a^2$, and subtract the value at our lower limit, $\infty$. One divided by infinity is exactly zero, so that term completely vanishes!

We started with a complex field and found a smooth, elegant potential function. Interestingly, this exact mathematical form appears when you calculate the potential on the axis of a uniform mass ring! Physics is full of these beautiful, recurring patterns.