The gravitational field due to a mass distribution is given by \(E =\frac{k}{x^3}\hat{i}\). Then the magnitude of gravitational potential at a distance 'd' along the x axis is-

The correct answer is option 2) i.e. \(\frac{k}{2d^2}\)

CONCEPT:​​​

• Gravitational Potential Energy: It is the energy possessed by a body at a certain point when work is done by the force of gravity in bringing the object from infinity to that point.

The gravitational potential energy between two masses m1 and m2 separated by a distance r is given by:

\(U = -\frac{Gm_1m_2}{r}\)​

EXPLANATION:

• The gravitational field (E) is the gravitational force per unit mass (F/m) that would be exerted on a small mass at that point. 
• The gravitational potential (V) at a point in a gravitational field is the work done per unit mass in bringing an object from infinity to that point. 

\(\Rightarrow V = \int \frac{F}{m}.dr = \int E.dr\)

Given that:

\(E =\frac{k}{x^3}\hat{i}\)

The magnitude of the gravitational potential at a distance 'd' along the x-axis:

\(\Rightarrow V = \int E.dx =\int _{d} ^{\infty} \frac{k}{x^3} dx\)

\(\Rightarrow V =[\frac{-k}{2x^2}] _d ^\infty \)

\(\Rightarrow V =\frac{-k}{2d^2}\)

\(\Rightarrow| V |=\frac{k}{2d^2}\)