Answer:
Option C
Explanation:
According to the question, the space between the two large parallel plates is filled with a material of uniform charge density $\rho$ i.e, it is a homogeneous medium , therefore the potential at any point between these plates is calculated by Poisson's equation as given below,
$\triangledown ^{2} V= -\frac{\rho}{\epsilon _{0}}$ ..........(i)
Since, we have to calculate the potential at any point along x- direction , hence the derivative of the potential along y and z directions is zero.
Hence, from Eq.(i) , we get
$\frac{d^{2} V}{dx^{2}}= -\frac{\rho}{\epsilon _{0}}$
Integreate on both the sides, we get
$\Rightarrow$ $\int \frac{d^{2} V}{dx^{2}}dx= -\int \frac{\rho}{\epsilon _{0}}dx\Rightarrow \frac{d V}{dx}=\frac{-\rho x}{\epsilon _{0}}-A$
Integrate both sides again, we get
$\Rightarrow$ $\int \frac{d^{} V}{dx^{}}dx= \int \frac{-\rho x}{\epsilon _{0}}dx-\int A dx$
$V= \frac{-\rho x^{2}}{2\epsilon _{0}}- A x-B$
$V=- \left(\frac{+\rho x^{2}}{2\epsilon _{0}}+ A x+B\right]$
