Two large parallel plates are located in vacuum. One of them serves as a cathode, a source of electrons whose initial velocity is negligible. An electron flow directed toward the opposite plates produce a space charge causing the potential in the gap between the plates to vary as \[V = a{x^{\dfrac{4}{3}}}.\] where, a is a positive constant and $x$ is the distance from the cathode.

The electric field between the plates as a function of $x$ is:
A. $E = a{x^{\dfrac{1}{3}}}$
B. $E = - a{x^{\dfrac{2}{3}}}$
C. $E = \dfrac{{ - 4a}}{3}{x^{\dfrac{1}{3}}}$
D. None

Hint: In order to solve this question, we should know about electric potential and electric field. Electric potential is the amount of work done while moving a unit test charge from one point to another and electric field is the field around a charge where any other electric charge can experience electric force. Here, we will use the general relation between electric potential and electric field to calculate the electric field as a function of x between the plates.

Formula used:
If $V$ is the electric potential as a function of variable $x$ written as $V(x)$ in the region of space, then electric field in that region is calculated as,
$E(x) = - \dfrac{{dV(x)}}{{dx}}$
Which is the derivative of potential function with respect to variable $x$.

Complete step by step answer:
According to the question, we have simply given that, the potential between the plates as a function of variable $x$.
\[V = a{x^{\dfrac{4}{3}}}\]
Now, in order to find electric field in this region we use the formula as,
$E(x) = - \dfrac{{dV(x)}}{{dx}}$
Where, \[V(x) = a{x^{\dfrac{4}{3}}}\].
Finding the derivative as
$E(x) = - \dfrac{{d(a{x^{\dfrac{4}{3}}})}}{{dx}}$
As we know, a is a constant so it will not be differentiated and $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ so,
$E(x) = - a\dfrac{{d({x^{\dfrac{4}{3}}})}}{{dx}}$
$\Rightarrow E(x) = - a\dfrac{4}{3}{x^{\dfrac{1}{3}}}$
$\therefore E(x) = \dfrac{{ - 4a}}{3}{x^{\dfrac{1}{3}}}$

Hence, the correct option is C.

Note: It should be remembered that, in the electric potential and electric field relation $E(x) = - \dfrac{{dV(x)}}{{dx}}$ the negative sign indicates that in the direction of away from the source of electric field in the region the electric potential keeps on decreasing so always keep in mind this negative sign while solving such questions.