Question

If gravitational potential at a place is $V = \left( {4xy - 2yz + 4xyz} \right)J/kg$ then find gravitational field intensity at potential $\left( {1m,1m,1m} \right)$
A. $9N/kg$
B. $\sqrt {104} N/kg$
C. $\sqrt {102} N/kg$
D. $10N/kg$

Answer 1

Gravitational potential is different as the work done per unit mass to take it away from the source point to infinity against the force of gravitation.
Gravitational potential, $V = \dfrac{W}{m}$ where W = work done in J and m = mass of the object in kg.

Complete step-by-step answer:
Whenever there is a concept of field in Physics, it represents the sphere of influence around an object with a property of either mass or charge, which exhibits an interaction in a profound way when a similar object is present in its field.
In the gravitational field, by the virtue of its mass, the body exerts its sphere of influence around itself wherein it exerts force on another body in the field which is proportional to the distance of separation of this body from the mass m.
The gravitational field intensity represents the sphere of influence of the object where it will exert a gravitational pull on another unit mass at a distance.
If F is the force applied on the object of mass m, the gravitational field intensity –
$E = \dfrac{F}{m}$
Also, $V = \dfrac{W}{m}$
The work done, $W = F \times s$ where s is the distance at which the object is present.
Substituting, we get –
$V = \dfrac{W}{m} \\\ V = \dfrac{{F \times s}}{m} \\\ rearranging, \\\ V = \dfrac{F}{m} \times s \\\ \therefore V = E \times s \\\$
Also, $E = \dfrac{V}{s}$
Since, in the question, there is a function, we need to take the differential.
$E = \dfrac{{dV}}{{ds}}$
Since, the function $V(x,y,z)$ has two variables and the field intensity is a vector quantity, we can write the above equation as :
$\overrightarrow E = - \dfrac{{\delta V}}{{\delta x}}i - \dfrac{{\delta V}}{{\delta y}}j - \dfrac{{\delta V}}{{\delta z}}k$
$\dfrac{{\delta V}}{{\delta x}} = \dfrac{\delta }{{\delta x}}\left( {4xy - 2yz + 4xyz} \right) = 4y - 0 + 4yz = 4y\left( {1 + z} \right)$

$\dfrac{{\delta V}}{{\delta y}} = \dfrac{\delta }{{\delta y}}\left( {4xy - 2yz + 4xyz} \right) = 4x - 2z + 4xz$
$\dfrac{{\delta V}}{{\delta z}} = \dfrac{\delta }{{\delta z}}\left( {4xy - 2yz + 4xyz} \right) = 0 - 2y + 4xy = 4xy - 2y$
Thus, the equation for the field intensity,
$\overrightarrow E = - 4y\left( {1 + z} \right)i - \left( {4x - 2z + 4xz} \right)j - \left( {4xy - 2y} \right)k$
Substituting the value of the coordinates of $\left( {x,y,z} \right) = \left( {1,1,1} \right)$, we get –
$\overrightarrow E = - 4\left( {1 + 1} \right)i - \left( {4 - 2 + 4} \right)j - \left( {4 - 2} \right)k \\\ \to \overrightarrow E = - 8i - 6j - 2k \\\$
By calculating the magnitude of the vector, we get –
$\overrightarrow {\left| E \right|} = \sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 2} \right)}^2}} \\\ \overrightarrow {\left| E \right|} = \sqrt {64 + 36 + 4} = \sqrt {104} J/kg \\\$

Therefore, the correct option is Option B.

Note: There is an electric-mechanical analogy to the electricity and gravitational field. These analogous terms are used to correlate and understand these phenomena and vice-versa. Here is the list of analogous terms.

Electrical	Mechanical
Charge, Q	Mass, M
Electric Field, E	Gravitation field, E
Electric potential, V	Gravitational potential, V
Coulomb’s law,$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}N$	Newton’s law,$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}N$
The constant $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N - {m^2}/{C^2}$	The constant $G = 6.67 \times {10^{ - 11}}{m^3}/kg - {\sec ^2}$