Question

What is the negative gradient of potential?
(A) Electric force
(B) Torque
(C) Electric current
(D) Electric field intensity

Answer 1

Hint: Electric field lines show us the direction of the electric field. The direction of the electric field is taken to be the direction of the force it would exert on the positive test charge. It is radially outwards towards positive charge and radially inwards to negative charge.

Complete step by step solution:
Electric potential can be defined as the amount of work done to bring the unit positive charge from infinity to that particular point.
We know that electric field lines go from positive charge to negative charge. This is the point where electric potential energy decreases. As we also know, electric field lines show us the direction of electric fields. So, the electric field points out in the direction in which the potential energy function decreases.
We know that, electric potential energy at any point due to the point charge $ q $ is given by –
 $ V = \dfrac{{kq}}{r} $
where, $ k $ is the constant of proportionality
Now, taking del operator on both sides of the above equation –
 $
  \nabla V = kq\nabla \left( {\dfrac{1}{r}} \right) \\
   \Rightarrow \nabla V = kq\left( {\dfrac{{\hat r}}{{{r^2}}}} \right) \\
   \Rightarrow - \nabla V = k\dfrac{q}{{{r^2}}}\hat r \cdots \left( 1 \right) \\
  $
We also know that –
 $ \vec E = k\dfrac{q}{{{r^2}}}\hat r \cdots \left( 2 \right) $
Hence,
 $ \vec E = - \nabla V $
So, mathematically, we can prove that the electric field is the negative gradient of the electric potential.
Hence, the correct option is (D).

Note:
The potential is positive just in case of positive charges as a result of we've to try and do work to maneuver a check charge from eternity thereto point. Currently if you calculate gradient of potential it'll offer the direction of most increase of potential however the potential will increase if we tend to go towards the central charge because 2 positive charges repel one another so direction of force ought to be within the direction of the potential.