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In a gravitational field, the gravitational potential is given by V=−KxV=-\frac{K}{x} J/kg. The gravitational field intensity at the point (2,0,3)(2,0,3) m is:

A

+K2+\frac{K}{2}

B

−K2-\frac{K}{2}

C

−K4-\frac{K}{4}

D

+K4+\frac{K}{4}

Step-by-Step Solution

  1. Relationship between Field and Potential: The gravitational field intensity (II or EE) is the negative gradient of the gravitational potential (VV). In one dimension (since VV depends only on xx), the relationship is given by: I=−dVdxI = -\frac{dV}{dx} (Reference: Concept of conservative forces, F=−dV/dxF = -dV/dx, applied to unit mass ).
  2. Differentiation: Given V=−Kx−1V = -Kx^{-1}. Differentiating with respect to xx: dVdx=ddx(−Kx−1)=−K(−1)x−2=Kx2\frac{dV}{dx} = \frac{d}{dx}(-Kx^{-1}) = -K(-1)x^{-2} = \frac{K}{x^2}
  3. Calculation: Substitute the derivative back into the field equation: I=−(Kx2)=−Kx2I = -\left( \frac{K}{x^2} \right) = -\frac{K}{x^2}
  4. Substitution: At the point (2,0,3)(2, 0, 3), the value of xx is 22. Substitute x=2x=2: I=−K22=−K4I = -\frac{K}{2^2} = -\frac{K}{4} The gravitational field intensity is −K4-\frac{K}{4} N/kg (or J/kg/m).
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