The gravitational field due to a mass distribution is `E=(A)/(x^(2))` in x-direction. Here, A is a constant, Taking the gravitational potential to be zero at infinity, potential at x is

Options

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

Answer

To find the gravitational potential \( V \) at a distance \( x \) from a mass distribution where the gravitational field \( E \) is given by \( E = \frac{A}{x^2} \), we can follow these steps:

### Step 1: Understand the relationship between gravitational field and potential
The gravitational potential \( V \) at a point is related to the gravitational field \( E \) by the equation:
\[ V = -\int E \, dx \]
This integral is evaluated from a reference point (in this case, infinity) to the point \( x \).

### Step 2: Set up the integral
Since the gravitational potential is taken to be zero at infinity, we need to evaluate the integral from infinity to \( x \):
\[ V(x) = -\int_{\infty}^{x} E \, dx \]
Substituting the expression for \( E \):
\[ V(x) = -\int_{\infty}^{x} \frac{A}{x^2} \, dx \]

### Step 3: Change the variable in the integral
To avoid confusion with the limits, we can change the variable of integration. Let’s denote the variable of integration as \( u \) instead of \( x \):
\[ V(x) = -\int_{\infty}^{x} \frac{A}{u^2} \, du \]

### Step 4: Evaluate the integral
Now we can evaluate the integral:
\[ V(x) = -A \left[ -\frac{1}{u} \right]_{\infty}^{x} = -A \left( -\frac{1}{x} + 0 \right) = \frac{A}{x} \]

### Step 5: Final expression for gravitational potential
Thus, the gravitational potential \( V \) at a distance \( x \) is given by:
\[ V(x) = -\frac{A}{x} \]

### Summary
The gravitational potential at a distance \( x \) from the mass distribution is:
\[ V(x) = -\frac{A}{x} \]