© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Relate the gravitational field to the potential
The gravitational field $E_x$ along the x-axis is related to the gravitational potential $V(x)$ by the relation:
$E_x = -\,\frac{dV}{dx}$
Step 2: Express the given information
We are given that
$E_x = \frac{A\,x}{\left(x^2 + a^2\right)^{3/2}}$
Substituting into $E_x = -\,\frac{dV}{dx}$ gives:
$-\frac{dV}{dx} = \frac{A\,x}{\left(x^2 + a^2\right)^{3/2}}$
Step 3: Set up the integral
Rearranging, we get:
$\frac{dV}{dx} = -\,\frac{A\,x}{\left(x^2 + a^2\right)^{3/2}}$
To find $V(x)$, integrate both sides with respect to $x$:
$\int dV = -\,A \,\int \frac{x}{\left(x^2 + a^2\right)^{3/2}} \,dx$
Step 4: Perform the integration
Consider the integral:
$\int \frac{x}{\left(x^2 + a^2\right)^{3/2}} \,dx$
A suitable substitution is $u = x^2 + a^2$. Then, $du = 2x \,dx$, which implies $x \,dx = \frac{du}{2}$. Substituting:
$\int \frac{x}{\left(x^2 + a^2\right)^{3/2}} \,dx
= \int \frac{1}{2} \frac{1}{u^{3/2}} \,du
= \frac{1}{2} \int u^{-3/2} \,du
$
The integral of $u^{-3/2}$ is:
$\int u^{-3/2} \,du = \int u^{-\frac{3}{2}} \,du = -\,2\,u^{-1/2}$
Hence,
$\int \frac{x}{\left(x^2 + a^2\right)^{3/2}} \,dx
= \frac{1}{2}(-\,2\,u^{-1/2}) + C
= -\,u^{-1/2} + C
$
Returning to the original variable $x$ (since $u = x^2 + a^2$):
$= -\frac{1}{\sqrt{x^2 + a^2}} + C
$
Step 5: Apply the boundary condition
We have:
$V(x) = -\,A \left(-\frac{1}{\sqrt{x^2 + a^2}}\right) + K
= \frac{A}{\sqrt{x^2 + a^2}} + K
$
To determine the constant $K$, we use the given condition that $V(x)\to 0$ as $x \to \infty$. As $x \to \infty$, $\frac{1}{\sqrt{x^2 + a^2}} \to 0$. Thus we require $V(\infty) = 0$ which implies $K = 0$. Therefore,
$V(x) = \frac{A}{\sqrt{x^2 + a^2}}.
$
Final Answer
$\displaystyle V(x) = \frac{A}{\sqrt{x^2 + a^2}}.$
