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 can follow these steps:

### Step 1: Understand the relationship between gravitational field intensity and potential
The gravitational field intensity \( E \) is related to the potential \( V \) by the equation:
\[ E = -\frac{dV}{dx} \]
This means that the gravitational field intensity is the negative gradient of the potential.

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

### Step 3: Integrate to find the potential
To find the potential \( V(x) \), we need to integrate \( dV \) from infinity to \( x \):
\[ V(x) = -\int_{\infty}^{x} \frac{Ax}{(A^2 + x^2)^{3/2}} \, dx \]

### Step 4: Change of variable for integration
To solve the integral, we can use a substitution. Let:
\[ x = A \tan(t) \quad \Rightarrow \quad dx = A \sec^2(t) \, dt \]
Then, we have:
\[ A^2 + x^2 = A^2 + A^2 \tan^2(t) = A^2 (1 + \tan^2(t)) = A^2 \sec^2(t) \]
Thus, the integral becomes:
\[ V(x) = -\int_{\infty}^{t_x} \frac{A^2 \tan(t)}{(A^2 \sec^2(t))^{3/2}} A \sec^2(t) \, dt \]
where \( t_x \) corresponds to \( x = A \tan(t) \).

### Step 5: Simplify the integral
The expression simplifies to:
\[ V(x) = -\int_{\infty}^{t_x} \frac{A^2 \tan(t) A \sec^2(t)}{A^3 \sec^3(t)} \, dt = -\int_{\infty}^{t_x} \frac{\tan(t)}{A} \, dt \]
This leads to:
\[ V(x) = -\frac{1}{A} \int_{\infty}^{t_x} \tan(t) \, dt \]

### Step 6: Evaluate the integral
The integral of \( \tan(t) \) is:
\[ -\ln|\cos(t)| + C \]
Evaluating from \( \infty \) to \( t_x \):
\[ V(x) = -\frac{1}{A} \left[ -\ln|\cos(t_x)| + 0 \right] = \frac{1}{A} \ln|\cos(t_x)| \]

### Step 7: Convert back to \( x \)
Since \( \tan(t_x) = \frac{x}{A} \), we have:
\[ \cos(t_x) = \frac{A}{\sqrt{A^2 + x^2}} \]
Thus:
\[ V(x) = \frac{1}{A} \ln\left(\frac{A}{\sqrt{A^2 + x^2}}\right) \]
This can be simplified to:
\[ V(x) = \frac{1}{A} \left( \ln(A) - \frac{1}{2} \ln(A^2 + x^2) \right) \]

### Final Result
The potential at point \( x \) is given by:
\[ V(x) = \frac{1}{A} \ln(A) - \frac{1}{2A} \ln(A^2 + x^2) \]