Q25P - Introduction to Electrodynamics

Question

Using Eqs. 2.27 and 2.30, find the potential at a distance z above the

center of the charge distributions in Fig. 2.34. In each case, compute $\mathit{E}\mathbf{ }\mathbf{⃗}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{ }\mathbf{⃗}\mathit{V}$ , and compare your answers with Ex. 2.1, Ex. 2.2, and Prob. 2.6, respectively. Suppose that we changed the right-hand charge in Fig. 2.34a to -q; what then is the potential at P? What field does that suggest? Compare your answer to Pro b. 2.2, and explain carefully any discrepancy.

Step-by-Step Solution

Answer

(a) The potential at a distance  $z$ from two point charges  $q$ separated by a distance  is  $\frac{1}{4\pi {\epsilon }_0}\frac{2q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}}$ and the electric field is $\frac{1}{4\pi {\epsilon }_0}\frac{2qz}{{\left(z^2+{\left(\frac{d}{2}\right)}^2\right)}^{3/2}}\hat{z}$ .

(b) The potential at a distance $z$  from a uniform line charge density $\lambda$  of length l is $\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{L+\sqrt{Z^2+L^2}}{-L+\sqrt{Z^2+L^2}}\right)$  and the electric field is $\frac{L\lambda }{2\pi {\epsilon }_0}\frac{1}{Z\sqrt{Z^2+L^2}}\hat{Z}$ .

(c) The potential at a distance $z$  from a uniform circular surface charge density $\sigma$ of radius $R\text{ is }\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right)$ and the electric field is $\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)\hat{z}$ .

Given data

(a) Two point charges  $q$ separated by a distance $d$ .

(b) A uniform line charge density $\lambda$  of length $2L$ .  

(c) A uniform circular surface charge density  $\sigma$ of radius $R$ .  

Potential and electric field

The potential at a distance $r$  from a charge $q$  is

  $\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here,  ${\epsilon }_0$ is the permittivity of free space.

$\mathbf{E}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{\nabla }}\mathbf{V}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\dots }\mathbf{\dots }\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The electric field as a function of potential is

 

Potential and electric field for part (a)

From equation (1), the potential at P from the two charges as shown in Fig. (a) is

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{2q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}} \end{gathered}$$

From equation (2), the electric field at P is

$$\begin{gathered} \overrightarrow{E}=-\overrightarrow{\nabla }\left(\frac{1}{4\pi {\epsilon }_0}\frac{2q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}}\right) \\ =-\frac{\partial }{\partial z}\left(\frac{1}{4\pi {\epsilon }_0}\frac{2q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}}\right)\hat{z} \\ =-\frac{2q}{4\pi {\epsilon }_0}\left(-\frac{1}{2}\right)\frac{2z}{{\left(z^2+{\left(\frac{d}{2}\right)}^2\right)}^{3/2}}\hat{z} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{2qz}{{\left(z^2+{\left(\frac{d}{2}\right)}^2\right)}^{3/2}}\hat{z} \end{gathered}$$

Thus, the potential is $\frac{1}{4\pi {\epsilon }_0}\frac{2q}{\sqrt{z^2+{\left(\frac{d}{2}\right)}^2}}$ and the electric field is $\frac{1}{4\pi {\epsilon }_0}\frac{2qz}{{\left(z^2+{\left(\frac{d}{2}\right)}^2\right)}^{3/2}}\hat{z}$ . If one of the point charges is changed to $-q$ then from the above calculation, the potential and the field is zero. This is wrong and only the $z$ component of the field will be zero. It will have non zero $x$ and $y$ components.

Potential and electric field for part (b)

From equation (1), the potential at P from a small length element  $dx$ on the length charge at a distance  $x$ from the mid point is

 $dV=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dx}{\sqrt{z^2+x^2}}$ 

The expression for the net field at P is

$V={\int }_{-L}^{L}dV$  

Substitute the value in the above equation and get

$$\begin{gathered} V=\frac{\lambda }{4\pi {\epsilon }_0}{\int }_{-L}^L\frac{dx}{\sqrt{z^2+x^2}} \\ =\frac{\lambda }{4\pi {\epsilon }_0}{\left[\ln \left(x+\sqrt{z^2+x^2}\right)\right]}_{-L}^L \\ =\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{L+\sqrt{Z^2+L^2}}{-L+\sqrt{Z^2+L^2}}\right) \end{gathered}$$

From equation (2), the electric field at P is

$$\begin{gathered} \overrightarrow{E}=-\overrightarrow{\nabla }\left[\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{L+\sqrt{z^2+L^2}}{-L+\sqrt{z^2+L^2}}\right)\right] \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\frac{\partial }{\partial \mathrm{z}}\ln \left(\frac{L+\sqrt{{\mathrm{z}}^2+L^2}}{-L+\sqrt{{\mathrm{z}}^2+L^2}}\right)\hat{z} \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\frac{\partial }{\partial \mathrm{z}}\ln \left(\frac{L+\sqrt{{\mathrm{z}}^2+L^2}}{-L+\sqrt{{\mathrm{z}}^2+L^2}}\right)\hat{z}=-\frac{\lambda }{4\pi {\epsilon }_0}\left[\frac{1}{L+\sqrt{z^2+L^2}}\frac{1}{2\sqrt{z^2+L^2}}2z-\frac{1}{-L+\sqrt{z^2+L^2}}\frac{1}{2\sqrt{z^2+L^2}}2z\right]\hat{z} \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\frac{Z}{\sqrt{Z^2+L^2}}\frac{-2L}{Z^2}\hat{Z} \end{gathered}$$

$$\begin{gathered} E=-\overrightarrow{\nabla }\left[\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{L+\sqrt{Z^2+L^2}}{-L+\sqrt{z^2+L^2}}\right)\right] \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\frac{\partial }{\partial z}\ln \left(\frac{L+\sqrt{z^2+L^2}}{-L+\sqrt{z^2+L^2}}\right)\hat{z} \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\left[\frac{1}{L+\sqrt{z^2+L^2}}\frac{1}{2\sqrt{z^2+L^2}}2z-\frac{1}{-L+\sqrt{z^2+L^2}}\frac{1}{2\sqrt{z^2+L^2}}2z\right]\hat{z} \\ =-\frac{\lambda }{4\pi {\epsilon }_0}\frac{2}{\sqrt{Z^2+L^2}}\frac{-2L}{Z^2}\hat{Z} \end{gathered}$$

Thus, the potential is $\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{L+\sqrt{z^2+L^2}}{-L+\sqrt{z^2+L^2}}\right)$and the electric field is $\frac{L\lambda }{2\pi {\epsilon }_0}\frac{1}{Z\sqrt{Z^2+L^2}}\hat{Z}.$

Potential and electric field for part (c)

From equation (1), the potential at P from a small circular ring of width $dr$  on the surface charge at a distance $r$  from the center is

$dV=\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 2\pi rdr}{\sqrt{z^2+r^2}}$

The expression for the net field at P is

 $V={\int }_0^{R}dV$ 

Substitute the value in the above equation and get

$$\begin{gathered} V=\frac{\sigma \pi }{4\pi {\epsilon }_0}{\int }_0^R\frac{2rdr}{\sqrt{z^2+r^2}} \\ =\frac{\sigma }{4{\epsilon }_0}{\left[2\sqrt{z^2+r^2}\right]}_0^R \\ =\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right) \end{gathered}$$

From equation (2), the electric field at P is

$$\begin{gathered} E=-\overrightarrow{\nabla }\left[\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right)\right] \\ =-\frac{\sigma }{2{\epsilon }_0}\frac{\partial }{\partial z}\left(\sqrt{z^2+R^2}-z\right)\hat{z} \\ =-\frac{\sigma }{2{\epsilon }_0}\left(\frac{1}{2\sqrt{z^2+R^2}}2z-1\right)\hat{z} \\ =\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)\hat{z} \end{gathered}$$

Thus, the potential is $\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right)$and the electric field is $\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)\hat{z}$