On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by Ax/( x² + a² )^(3/2) in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

Solution

Physics sir · 4 steps

Potential is the integral of the field, and the substitution u = x² + a² does all the work here.

1. Field to potential. With V = 0 at infinity, the potential is minus the integral of the field from infinity up to the point x.

   V(x) = -∫_∞ˣ Ax/(x² + a²)^(3/2) dx

2. Substitute. Put u = x² + a², so x dx = 1/2du. The integrand becomes A/2u^(-3/2), which integrates to A/2 × u^(-1/2)/(-1/2) = -A/√u.
3. Apply the limits. At infinity, -A/√u goes to 0. That leaves V(x) = -(-A/√(x² + a²) - 0) = A/√(x² + a²).
4. So. The magnitude of the potential is A/(x² + a²)^(1/2), option (c). As a check, differentiating it gives Ax/(x² + a²)^(3/2) in magnitude, which is the field you were given.

The idea to keep. Field is minus the slope of potential, so potential is minus the integral of field from the zero point. When you see x multiplying a power of (x² + a²), substitute u = x² + a².

Answer from the NTA final answer key. Solution by PrepWiser.