Question

A non-uniform linear charge distribution given by $\mathit{\lambda }\mathbf{=}\mathit{b}\mathit{x}$, where bis a constant, is located along an xaxis from x = 0to x = 0.20m. If b = 20nC/m²and V = 0at infinity, what is the electric potential at (a) the origin and (b) the point y = 0.15mon the yaxis?

Short answer

1. The electric potential at the origin on the y-axis is 36 V.
2. The electric potential at the point on the y-axis is 18 V.

Step-by-step solution

The given data

1. Non-uniform linear charge is $\lambda =bx$ where b = 20nC/m², that is located along an x-axis from x = 0 to x = 0.2m.
2. The distance of the point, y = 0.15m
3. The value of the electric potential is V = 0 at infinity.

Understanding the concept of the electric potential

At first, we can get the value of the charge of the particle using the concept of linear density. Then this value is substituted in the formula of the electric potential considering the given distances of both the cases as given, and then this determines the value of the potentials.

Formulae:

The electric potential at the point due to charge,$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$ (i)

The linear charge density of a distribution, $\lambda =\frac{q}{l}$ (ii)

a) Calculation of the electric potential

The charge due to non-uniform linear charge distribution can be given using equation (ii) as follows:

$\begin{array}{rcl}q& =& \lambda l\\ & =& \left(bx\right)x\\ & =& b{x}^2\end{array}$(substituting the given values)

The electric potential at the origin due to the charge is given using equation (i) and the given data as:

$\begin{array}{rcl}V& =& \frac{1}{4\pi {\epsilon }_0}\frac{b{x}^2}{x}\\ & =& 9\times {10}^9\times \frac{20\times {10}^{-9}\times {\left(0.2\right)}^2}{\left(0.2\right)}\\ & =& 36V\end{array}$

Hence, the value of the potential is 36V.

b) Calculation of the electric potential at the point

At d = 0.15m on the y-axis, the potential energy at the axial point is given using equation (i) as follows:

$\begin{array}{rcl}V& =& \frac{1}{4\pi {\epsilon }_0}\underset{0}{\overset{0.20}{\int }}\frac{bx}{\sqrt{x^2+d^2}}dx\\ & =& {\left[\frac{b}{4\pi {\epsilon }_0}\left(\sqrt{x^2+d^2}\right)\right]}_0^{0.20}\\ & =& \left(20\times {10}^{-9}\times 9\times {10}^9\right)\left[\left(\sqrt{0.{20}^2+0.{15}^2}\right)-\left(\sqrt{0.{15}^2}\right)\right]\\ & =& 18V\end{array}$

Hence, the value of the potential is 18V.