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$\newcommand{\sech}{\rm sech}$
We give several examples to explain how to obain the nature of energy spectrum without solving the problem completely.
$\newcommand{\Label}[1]{\label{#1}}\newcommand{\Prime}{{^\prime}}\newcommand{\pp}[2][]{\frac{\partial#1}{\partial #2}}\newcommand{\PP}[2][]{\frac{\partial^2#1}{\partial #2^2}}\newcommand{\dd}[2][]{\frac{d#1}{d #2}}\newcommand{\DD}[2][]{\frac{d^2#1}{d #2^2}}$
A body falling under the earth's gravitational field experiences a side ways deflection due to Coriolis force. An expression for this deflection is obtained and numerical values are estimated.
Question
A conical surface ( an empty ice cream cone ) carries a uniform surface charge density $\sigma_0$. The height of the cone and the radius of the base are both equal to $a$. Find the potentials at the vertex of the cone and at the center of the base.
Solution
We divide the conical surface into thin rings by planes parallel to the base. Let $2 \alpha$ be the angle of the cone.
\FigNoNumX{50,10}{60}{60}{MidSemAnsQ1}{}
Charge on the annular ring between $x$ and $x+dx=2\pi
xdx\sigma$
\begin{align*}
d\phi=&\frac{1}{4\pi\epsilon_0}~\frac{2\pi
xdx\sigma}{\sqrt{x^2+r^2}}\\
\phi=&\int_0^a \frac{1}{4\pi\epsilon_0}~\frac{2\pi xdx\sigma}
{\sqrt{x^2+r^2}}\\
=&\frac{\sigma}{2\epsilon_0}\int_0^a\frac{xdx}{\sqrt{x^2+a^2}}\\
=&\frac{\sigma}{2\epsilon_0}\left[\sqrt{a^2+r^2}-r\right]\\
\sqrt{a^2+r^2} = & r(1+a^2/r^2)^{1/2} \approx r + \frac{a^2}{2r}
\end{align*}
For large distances $r\gg a$
\begin{align*}
Consider a ring shaped surface of the shell between $x$
and $x+dx$.\\
Radius of sphere $=a$\\
Radius of the ring $=a\sin\theta$\\
$x=a\cos\theta$\\
Area of the ring =$(ad\theta)\times a\sin\theta\times
2\pi$\qquad\quad $dx=a\sec^2\theta d\theta$\\
Charges density $=\frac{Q}{4\pi a^2}=\sigma$
\begin{align*}
\therefore~d\phi=&\left(\frac{1}{4\pi\epsilon_0}\right)
\frac{\sigma\times2\pi a^2\sin\theta d\theta}{R}\qquad\qquad\qquad{QP=r-x}\\
=&\left({1}{4\pi\epsilon_0}\right)\left({Q}{2}\right)
The potential due to a system of point charges \(q_k\) at position \(\vec{r}_k\) is given by
Let a point \(P\) be at a distance \(r\) from the center of the spherical shell;
Consider a ring shaped surface of the shell between \(x\) and \(x+dx\)
Radius of the sphere = \(a\)
Radius of the ring =$a\sin\theta$
$x=a\cos\theta$
Area of the ring =$(ad\theta)\times(a\sin\theta)\times 2\pi$
$dx=a\sec^2\theta d\theta$
charge density $\sigma =\frac{Q}{4\pi a^2}$
This is very similar to the full sphere having uniform charged density. The thick shell is divided into thin shells lying between \(x\) and \(x+dx\). Potential at the given point P due to a thin shell of radius $x>R$ is$\phi_1$:
\begin{eqnarray}
\phi_1&=&\frac{4\pi x^2dx\rho}{4\pi\epsilon_0x}
=\frac{\rho xdx}{\epsilon_0} \label{eq34}
\end{eqnarray}
Therefore potential at a point \(r\), due to shells between \(a\) and \(b\), is $\phi_2$
Using the expression for energy density we can compute the energy stored in a charged capacitor. The electrostatic energy is given by the energy per unit volume, ,multiplied by the volume enclosed between {\it i.e.} \( \frac{\epsilon_0}{2}|\vec{E}|^2\) the plates. Thus we get
\begin{eqnarray}
W&=&\frac{\epsilon_0}{2}(Ad)(\frac{V_0}{d})^2\\ \label{eq19}
\nonumber&=&\frac{\epsilon_0A}{2d}(V_0)^2 \\
&=&\frac{Q^2}{2C} \label{eq20},
\end{eqnarray}
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