Question

The space between the two large parallel plates is filled with a material of uniform charge density $\rho$. Assume that one of the plate is kept at $x=0 .$ The potential at any point $x$ between these plates is given by $(A$ and $B$ are constants $)$.

• A. $-\frac{\rho x^{3}}{2 \epsilon_{0}}$
• B. $-\left(\frac{\rho x^{2}}{2 \epsilon_{0}}+A x\right)$
• C. $-\left(\frac{\rho x^{2}}{2 \epsilon_{0}}+A x+B\right)$
• D. $-\left(\frac{\rho x^{3}}{4 \epsilon_{0}}+A x^{2}+B x\right)$

Answer 1

Answer: (C)

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,
$\nabla^{2} V=-\frac{\rho}{\varepsilon_{0}}$
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
$\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\varepsilon_{0}}$
Integrate on both the sides, we get
$\Rightarrow \int \frac{d^{2} V}{d x^{2}} d x=-\int \frac{\rho}{\varepsilon_{0}} d x$
$ \Rightarrow \frac{d V}{d x}=\frac{-\rho x}{\varepsilon_{0}}-A$
Integrate on both sides again, we get
$ \Rightarrow \int \frac{d V}{d x} d x =\int \frac{-\rho x}{\varepsilon_{0}} d x-\int A d x $
$V =\frac{-\rho x^{2}}{2 \varepsilon_{0}}-A x-B $
$ \therefore V =-\left(\frac{+\rho x^{2}}{2 \varepsilon_{0}}+A x+B\right] $