- #1
Plantation
- 17
- 1
- TL;DR Summary
- Griffith, Electrodynamics, understanding Example 4.8.
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now.
> Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin.
In the page 196, in the first paragraph, the author argues as follows :
We could, of course, obtain the field of ##\sigma_b## by direct integration
$$ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \int ( \frac{\hat{\mathfrak{r}}}{\mathfrak{r}^2}) \sigma_b da$$ ( where ##\mathfrak{r}## is the separation vector. I still don't know how to write cursive ##r## in physics Forum ).
But, as in the case of the conducting plane, there is a nicer solution by the method of images.
Indeed, if we replace the dielectric by a single point charge ##q_b## at the position ##(0,0, -d)##, we have
$$ V = \frac{1}{4 \pi \epsilon_0}[ \frac{q}{\sqrt{x^2 +y^2 + (z-d)^2 }} + \frac{q_b}{\sqrt{x^2+ y^2+(z+d)^2}}]. \tag{4.52}$$
in the region ##z>0##. Meanwhile, a charge ##(q+q_b)## at ##(0,0.d)## yields the potential
$$ V= \frac{1}{4\pi \epsilon_0} [ \frac{q + q_b}{\sqrt{x^2+y^2+(z-d)^2}}] . \tag{4.53}$$
for the region ##z<0##.
Q.1. Why ##(4.52)## holds for region ##z>0## ? And why we consider a charge ##(q+q_b)## at ##(0,0,d)## and why the potential ##(4.53)## it yelds holds for the region ##z<0## ?
I've only skimmed this issue roughly, but I don't understand it correctly.
(Continuing arguemnt ) Taken together, Eqs. 4.52 and 4.53 constitute a function that satisfies Poisson' equation with a point charge ##q## at ##(0,0,d)## ( what exact form of equation is it? ), which goes to zero at infinity, which is continuous at the boundary ##z=0##, and whose normal derivative exhibits the discontinuity appropriate to a surface charge ##\sigma_b## at ##z=0## :
$$-\epsilon_0 (\frac{\partial V}{\partial z}|_{z=0^{+}}- \frac{\partial V}{\partial z}|_{z=0^{-}} ) = -\frac{1}{2 \pi}( \frac{\chi_e}{\chi_e +2}) \frac{qd}{(x^2+y^2+d^2)^{3/2}}. \tag{1}$$
Q.2. Can we really show that the ##(4.52)## and ##(4.53)## together satisfies the Poisson's equation? Is ##(1)## the result of a direct calculation obtained by substituting ##(4.52)## and ##(4.53)## ( and using ##(4.51)## in his book ) ? Why such result ##(1)## guarantees the associated boundary condition? Finally, how can we know that the boundary conditions Griffith arranges- goes to zero at infinity, continuous at the boundary ##z=0##, normal derivative's exhibition of the discontinuity appropriate to a surface charge ##\sigma_b## at ##z=0## .. - are 'exact' boundary conditions for solving the original problem ( Example 4.8 ), so that we may apply the method of images ( the uniqueness theorem ) ?
( Cont. ) Accordingly, this is the correct potential for our problem. In particular, the force on ##q## is :
$$ \mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{qq_b}{(2d)^2} \hat{\mathbf{z}} = - \frac{1}{4\pi \epsilon_0}(\frac{\chi_e}{\chi_e + 2})\frac{q^2}{4d^2}\hat{\mathbf{z}}. \tag{4.54} $$
Q.3. Why the first equality in ##(4.54)## is true? Can anyone hint? What should I plug into which formula? Please comment what I should recall.
P.s. Q.4. There is additional question ( may skip.. ) In the final paragraph in the solution, Griffith wrote as follows :
I don't understand the argument in the part highlighted in red. Could someone please explain it more friendly to me?
Can anyone teach me/ help? Thanks for reading.
> Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin.
In the page 196, in the first paragraph, the author argues as follows :
We could, of course, obtain the field of ##\sigma_b## by direct integration
$$ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \int ( \frac{\hat{\mathfrak{r}}}{\mathfrak{r}^2}) \sigma_b da$$ ( where ##\mathfrak{r}## is the separation vector. I still don't know how to write cursive ##r## in physics Forum ).
But, as in the case of the conducting plane, there is a nicer solution by the method of images.
Indeed, if we replace the dielectric by a single point charge ##q_b## at the position ##(0,0, -d)##, we have
$$ V = \frac{1}{4 \pi \epsilon_0}[ \frac{q}{\sqrt{x^2 +y^2 + (z-d)^2 }} + \frac{q_b}{\sqrt{x^2+ y^2+(z+d)^2}}]. \tag{4.52}$$
in the region ##z>0##. Meanwhile, a charge ##(q+q_b)## at ##(0,0.d)## yields the potential
$$ V= \frac{1}{4\pi \epsilon_0} [ \frac{q + q_b}{\sqrt{x^2+y^2+(z-d)^2}}] . \tag{4.53}$$
for the region ##z<0##.
Q.1. Why ##(4.52)## holds for region ##z>0## ? And why we consider a charge ##(q+q_b)## at ##(0,0,d)## and why the potential ##(4.53)## it yelds holds for the region ##z<0## ?
I've only skimmed this issue roughly, but I don't understand it correctly.
(Continuing arguemnt ) Taken together, Eqs. 4.52 and 4.53 constitute a function that satisfies Poisson' equation with a point charge ##q## at ##(0,0,d)## ( what exact form of equation is it? ), which goes to zero at infinity, which is continuous at the boundary ##z=0##, and whose normal derivative exhibits the discontinuity appropriate to a surface charge ##\sigma_b## at ##z=0## :
$$-\epsilon_0 (\frac{\partial V}{\partial z}|_{z=0^{+}}- \frac{\partial V}{\partial z}|_{z=0^{-}} ) = -\frac{1}{2 \pi}( \frac{\chi_e}{\chi_e +2}) \frac{qd}{(x^2+y^2+d^2)^{3/2}}. \tag{1}$$
Q.2. Can we really show that the ##(4.52)## and ##(4.53)## together satisfies the Poisson's equation? Is ##(1)## the result of a direct calculation obtained by substituting ##(4.52)## and ##(4.53)## ( and using ##(4.51)## in his book ) ? Why such result ##(1)## guarantees the associated boundary condition? Finally, how can we know that the boundary conditions Griffith arranges- goes to zero at infinity, continuous at the boundary ##z=0##, normal derivative's exhibition of the discontinuity appropriate to a surface charge ##\sigma_b## at ##z=0## .. - are 'exact' boundary conditions for solving the original problem ( Example 4.8 ), so that we may apply the method of images ( the uniqueness theorem ) ?
( Cont. ) Accordingly, this is the correct potential for our problem. In particular, the force on ##q## is :
$$ \mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{qq_b}{(2d)^2} \hat{\mathbf{z}} = - \frac{1}{4\pi \epsilon_0}(\frac{\chi_e}{\chi_e + 2})\frac{q^2}{4d^2}\hat{\mathbf{z}}. \tag{4.54} $$
Q.3. Why the first equality in ##(4.54)## is true? Can anyone hint? What should I plug into which formula? Please comment what I should recall.
P.s. Q.4. There is additional question ( may skip.. ) In the final paragraph in the solution, Griffith wrote as follows :
I don't understand the argument in the part highlighted in red. Could someone please explain it more friendly to me?
Can anyone teach me/ help? Thanks for reading.
