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. $-\frac{\rho x^3}{2{ϵ}_0}$
• B. $-\left(\frac{\rho x^2}{2{ϵ}_0}+Ax\right)$
• C. $-\left(\frac{\rho x^2}{2{ϵ}_0}+Ax+B\right)$
• D. $-\left(\frac{\rho x^3}{4{ϵ}_0}+A{x}^2+Bx\right)$

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,
${\nabla }^{2}V=-\frac{\rho }{{\epsilon }_0}$
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
$\frac{d^{2}V}{d{x}^2}=-\frac{\rho }{{\epsilon }_0}$
Integrate on both the sides, we get
$\Rightarrow \int \frac{d^{2}V}{d{x}^2}dx=-\int \frac{\rho }{{\epsilon }_0}dx$
$\Rightarrow \frac{dV}{dx}=\frac{-\rho x}{{\epsilon }_0}-A$
Integrate on both sides again, we get
$\Rightarrow \int \frac{dV}{dx}dx=\int \frac{-\rho x}{{\epsilon }_0}dx-\int Adx$
$V=\frac{-\rho x^2}{2{\epsilon }_0}-Ax-B$
$\therefore V=-\left(\frac{+\rho x^2}{2{\epsilon }_0}+Ax+B\right]$