The electric potential decreases uniformly from 120V to 80V as one moves on the X-axis from x = - 1cm to x = +1cm. The electric field at the origin
(a) Must be equal to 20V/cm
(b) May be equal to 20V/cm
(c) May be greater than 20V/cm
(d) May be less than 20V/cm

Hint:We are given change in the potential and we are moving from one point to another. We can find the electric field by using the relationship between electric field and electric potential. The electric potential is a scalar quantity while the electric field is a vector quantity. While calculating the electric field using the gradient formula, the distance is to be taken in meters and electric potential in Volts.

Complete step by step answer:
The change in electric potential as per the question$\Delta V=(80-120)V=-40V$
The change in the position, the initial coordinates are $(-1,0)$and the final coordinates are $(1,0)$. So, the change is $\Delta x=1-(-1)=2cm=0.02m$
We need to find the electric field. We know electric field can be given as negative gradient of the electric potential, thus we can write it as:$E=-\dfrac{\Delta V}{\Delta x}$
$\Rightarrow E=-\dfrac{\Delta V}{\Delta x}$
$\Rightarrow E=-\dfrac{(-40)}{0.02}$
$\therefore E=2000V/m$
Changing it into $V/cm$ we get $20V/cm$.

So, the correct options are B and C.

Note:Electric potential is a scalar quantity and it is assumed to be zero at the infinity. If we want to move a charge from one point to another and if the electric potential is same at both the points work done will be zero because no work is required to be done to move a charge on an equipotential surface.The electric potential is related to electric field and electric field is defined as the negative gradient of electric potential.