The gravitational field due to a mass distribution is `E=K//x^(3)` in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

Options

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

Answer

To find the gravitational potential \( V(x) \) at a distance \( x \) from a mass distribution, given that the gravitational field \( E \) is defined as:

\[ E = \frac{K}{x^3} \]

where \( K \) is a constant, we can follow these steps:

### Step 1: Understand the relationship between gravitational field and potential
The gravitational field \( E \) is related to the gravitational potential \( V \) by the equation:

\[ E = -\frac{dV}{dx} \]

This means that the gravitational potential can be found by integrating the gravitational field.

### Step 2: Set up the integral for potential
Since we know \( E \), we can express the change in potential \( dV \) as:

\[ dV = -E \, dx = -\frac{K}{x^3} \, dx \]

### Step 3: Integrate to find the potential
To find the potential at a distance \( x \) from infinity (where we take \( V(\infty) = 0 \)), we need to integrate from infinity to \( x \):

\[ V(x) - V(\infty) = -\int_{\infty}^{x} \frac{K}{r^3} \, dr \]

Since \( V(\infty) = 0 \), we have:

\[ V(x) = -\int_{\infty}^{x} \frac{K}{r^3} \, dr \]

### Step 4: Evaluate the integral
Now we compute the integral:

\[ V(x) = -\left[ -\frac{K}{2r^2} \right]_{\infty}^{x} \]

This simplifies to:

\[ V(x) = \left[ \frac{K}{2r^2} \right]_{\infty}^{x} \]

Evaluating the limits gives:

\[ V(x) = \frac{K}{2x^2} - \frac{K}{2(\infty)^2} \]

Since \( \frac{K}{2(\infty)^2} = 0 \), we find:

\[ V(x) = \frac{K}{2x^2} \]

### Final Answer
Thus, the gravitational potential at a distance \( x \) is:

\[ V(x) = \frac{K}{2x^2} \]

---