Gravitational field intensity is given by` E = Ax/((A^2 + x^2)^(3/2))` then find out potential at x (Assume potential at infinity = 0)

Options

A. `- 2A/(sqrt(A^2 + x^2)`
B. `- A/(sqrt(A^2 + x^2)`
C. `- A/3(sqrt(A^2 + x^2)`
D. `- 3A/(sqrt(A^2 + x^2)`

Answer

To find the gravitational potential \( V(x) \) at a point \( x \) given the gravitational field intensity \( E = \frac{Ax}{(A^2 + x^2)^{3/2}} \), we will follow these steps:

### Step 1: Understand the relationship between electric field and potential
The relationship between the electric field \( E \) and potential \( V \) is given by:
\[ E = -\frac{dV}{dx} \]
This means that the electric field is the negative gradient (rate of change) of the potential.

### Step 2: Write the expression for \( dV \)
From the relationship above, we can express \( dV \) as:
\[ dV = -E \, dx \]
Substituting the expression for \( E \):
\[ dV = -\frac{Ax}{(A^2 + x^2)^{3/2}} \, dx \]

### Step 3: Integrate to find potential
To find the potential \( V(x) \), we need to integrate \( dV \):
\[ V(x) = -\int \frac{Ax}{(A^2 + x^2)^{3/2}} \, dx \]
We will integrate this from \( x \) to \( \infty \) since the potential at infinity is given to be zero:
\[ V(x) = -\left[ \int_{\infty}^{x} \frac{Ax}{(A^2 + x^2)^{3/2}} \, dx \right] \]

### Step 4: Perform the integration
To solve the integral, we can use a substitution method. Let:
\[ u = A^2 + x^2 \implies du = 2x \, dx \implies dx = \frac{du}{2x} \]
When \( x = \infty \), \( u = \infty \) and when \( x = x \), \( u = A^2 + x^2 \).

Substituting in the integral:
\[ V(x) = -\int_{\infty}^{A^2 + x^2} \frac{A}{(u)^{3/2}} \cdot \frac{du}{2} \]
This simplifies to:
\[ V(x) = -\frac{A}{2} \left[ -\frac{2}{\sqrt{u}} \right]_{\infty}^{A^2 + x^2} \]
Evaluating the limits:
\[ V(x) = -\frac{A}{2} \left( 0 - \left( -\frac{2}{\sqrt{A^2 + x^2}} \right) \right) = \frac{A}{\sqrt{A^2 + x^2}} \]

### Step 5: Final expression for potential
Thus, the potential at a point \( x \) is given by:
\[ V(x) = \frac{A}{\sqrt{A^2 + x^2}} \]