Question

The electric potential decreases uniformly from V to –V along x-axis in a coordinate system as we move from a ($ - {x_0}$, 0) to (${x_0}$, 0), then the electric field at the origin.
(A) Must to equal to $\dfrac{V}{{{x_0}}}$;
(B) May be equal to $\dfrac{V}{{{x_0}}}$;
(C) Must be greater than $\dfrac{V}{{{x_0}}};$
(D) May be less than $\dfrac{V}{{{x_0}}}$;

Answer 1

Hint: To answer this question we should be knowing the concept of electric field. Once we develop the formula we have to put the values in the expression from the question. After that we have to evaluate the expression to get the required answer.

Complete step by step answer:
We should know that V decreases uniformly.
The formula of electric field:
$E = \dfrac{{ - \Delta V}}{{\Delta h}}$
Now we have to put the values in the above expression and evaluate:
$
  E = \dfrac{{ - (V - ( - V))}}{{( - {x_0} - {x_0})}} \\
   \Rightarrow E = \dfrac{{ - 2{V_0}}}{{ - 2{x_0}}} = \dfrac{{{V_0}}}{{{x_0}}} \\
 $

Hence we can say that the electric field at the origin Must be greater than $\dfrac{V}{{{x_0}}};$ so the correct answer is option C.

Note: We should know that electric potential is defined as the amount of work that is required to transfer a unit of charge from a specific reference point to a particular point inside the field, but in the absence of an acceleration. The unit of electric potential is volt or denoted by V.
It should also be known that the electric field is defined as the amount of electric force per unit charge. We should notice that the direction of the field that is taken is directed towards the force that would exert a positive test charge. The electric field is known to radially outward from a positive charge and then it is directed towards the radially transmitted negative point charge.
The formula of electric field is given by $E = kQ/{r^2}$ where k is the constant value.
We should not confuse between electric field and electric field intensity. Electric field intensity is defined as the strength of the electric field at any specific point. It is also known to be a vector quantity.
$E = kQ/{r^2}$