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# Let V and E denote the gravitational potential and gravitational field at a point, respectively. It is possible to have This question has multiple correct options(a) $V = 0$and $E = 0$(b) $V = 0$and $E \ne 0$ (c) $V \ne 0$and $E = 0$ (d)$V \ne 0$and $E \ne 0$

Last updated date: 20th Jun 2024
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Hint: Gravitational potential at a point is nothing but work done in bringing a unit mass from infinity separation to that particular point. Whereas, the gravitational field is the amount of force experienced by some mass placed in the gravitational potential at that point. Relation between field and potential is given by $E = - \dfrac{{dV}}{{dr}}$ where, E is field and V is potential and r is position variable.

Formula Used:
Relation between electric field and potential: $E = - \dfrac{{dV}}{{dr}}$ ……(1)
Where,
V is potential at the given point differentiated with respect to distance from the source.

Given:
1. Potential at a point: V
2. Electric field at a point: E

To find: The possible combinations of V and E.

Step 1:
In case the point at which the electric field and potential is being measured is at infinite distance from the source: $V = 0$. Putting $V = 0$in eq (1), we get $E = 0$. So, option (a) is correct.

Step 2:
In case the point lies inside a spherical shell of mass M and radius R, gravitational potential will be given by: $V = \dfrac{{GM}}{R}$, where G is gravitational constant. Here, the potential is constant as it does not depend on the distance from the source. Now, using eq (1):
$E = - \dfrac{{d(\dfrac{{GM}}{R})}}{{dr}} \\ E = 0 \\$
In this case, $V \ne 0$and $E = 0$, so option (c) is also correct.

Step 3:
In the general case, when V depends on r, its derivative(E) will also be a function of r. For example, Potential for a point outside a spherical shell located at a distance r from the centre will be given by: $V = \dfrac{{GM}}{r}$. Use eq (1) to find E:
$E = - \dfrac{{d(\dfrac{{GM}}{r})}}{{dr}} \\ E = \dfrac{{GM}}{{{r^2}}} \\$
Here, $V \ne 0$and $E \ne 0$, so option (d) is also correct.