An electric field \[\vec E = 20x\hat iN/C\] exists in the space. Take the potential at\[\left( {1m,2m} \right)\]to be zero. Find the potential at the origin:-
(1)\[2volt\]
(2) \[3volt\]
(3) \[6volt\]
(4) \[10volt\]

Hint: For this, we can use the relationship between the electric field vector, electric potential and displacement vector. As x and y coordinates are given in the question, by substituting the values, we can easily determine the electric field vector.

Formula used:
As we know the formula of potential,
\[dV = - \vec E.dx\]
Here E is the electric field intensity (that is given), dV is the change in potential. By using the potential limits and integrating the potential term, we can easily calculate the value of potential at the origin.

Complete step by step solution:
As we know, \[dV = - \vec E.dx\] ------- (1)
\[\vec E = 20x\hat iN/C\]
Substitute the value of electric field intensity in equation (1), we get-
\[dV = - 20x\hat i.dx\]
Now integrate this resulted equation by taking the limits, we get-
\[V = - 20{\left[ {\dfrac{{{x^2}}}{2}} \right]_1}^0\]
\[ 0,1\] (Upper and lower limits respectively) After substituting the upper and lower limits, we get-
\[\Rightarrow V = - 20\left[ {0 - \dfrac{1}{2}} \right] = 10volts\]
Hence the potential at the origin is \[10volts\].

So, the correct answer is “Option D”.

Note:
An electric field is the area surrounded by charge where the effect of charges can be measured. If the field is directed from lower potential to the higher potential then the direction will be positive and if the field is directed from higher to lower potential then the direction will be negative. This electric field exists only and only if there is a potential difference. If there is no change, means electric potential difference is uniform or equipotential surfaces.