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I've been trying to answer the following question but I'm stuck at one step. The question essentially states that a magician is trying to perform a "floating objects" act, for which she has a thin copper staff that is $L=1.5 \rm m$ long and places it on a metal plate which charges it uniformly with $25*10^{-9} \rm C$, he then lays it down horizontally and wants to place a small ball above the thin copper rod, right above one of the ends, and wants it to float $d=0.5 \rm m$ above that end of the rod. How much charge should the sphere (small ball) have to float at that height? The hint says start with electric potential, and then use either electric field or potential energy to answer the question.

My attempt to answer the question:

I started by finding the total electric potential on the ball using

$$ V(d)=\int \frac{K\,dq}{r} $$ where $dq=\lambda\,dL=\lambda dx, r=\sqrt{x^2+d^2}$ \begin{align} V(d)&=K*\lambda\int \frac{dx}{\sqrt{x^2+d^2}}\\&={K\,\dfrac{Q}{L}} [\ln(\sqrt{x^2+d^2}+x)] \end{align} evaluated from $x=0\,\rm m$ to $x=L$.

I can plug the numbers in and find total $V$, but I'm not sure how to proceed from there. I thought about using $V$ to find $E$ since $$ \Delta V=-\int \vec{E}\cdot d\vec{r} $$ and then use $$ F_e=Q\,E=mg\Rightarrow Q=\dfrac{mg}{E} $$ but the problem with this is I only have $V$ and not $\Delta V$, so I can't find $E$. I feel like I'm missing something obvious. I also tried relating electric potential energy with gravitational potential energy but couldn't quite figure out a way to relate the two. Any help would be greatly appreciated.

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    $\begingroup$ The problem description isn't quite clear to me. Please draw a diagram of the configuration of the ball and rod in the case the ball is floating. $\endgroup$ – Alfred Centauri Feb 5 '15 at 12:07
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Your derivation of $V$ is fine, the $E$ not so good. But for the potential you did something better. In fact, you computed that potential for any height $d$ above the rod. So compute $V$ with a $d=0$, then compute it with your actual $d$ and then use the two to get an electric potential difference. Use the charge of the small sphere and the electric potential difference to get a difference in electrical potential energy. Equate that to the difference in gravitational potential energy, $mg\Delta h=mgd$.

Now, if you really want to do the trick, I'd the small sphere above the center of the the rod so that it can be an unstable equilibrium instead of immediately being pushed away from the rod. To compute that case, just have $x$ go from $-L/2$ to $+L/2$ instead of $0m$ to $L$. Or imagine two rods of length $L/2$, so you end up with $4$ times the potential, so $1/4$ the charge needed on the small sphere. That ball will still fall, but only when bumped by the slightest disturbance.

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  • $\begingroup$ Interesting. I didn't think about equipotential lines. It makes sense. Thank you. $\endgroup$ – PoweredByOrange Feb 5 '15 at 19:19
  • $\begingroup$ One quick question, how could you equate electric potential energy to gravitational potential energy? What's the relation between the two? $\endgroup$ – PoweredByOrange Feb 5 '15 at 19:19
  • $\begingroup$ @PoweredByOrange Imagine placing the sphere near the rod, both are positively charged, so the sphere gets pushed up while gravity is pushing down. As you raise the height, the gravitational potential energy increases, but the electric potential energy decreases. Eventually the gradients of the two potential energies are equal and opposite, that's where the ball could rest (when in the center), since the forces balance. To see how high it jumps you want to have no kinetic energy, so that's where the two potential energy differences (from 0m to d) are equal and opposite, that's what I computed $\endgroup$ – Timaeus Feb 6 '15 at 7:14
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From http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml, the electric field is:

$$\begin{align}E_p&=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{b}+\frac{1}{L+b}\right)\\ &=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{0.5}+\frac{1}{2}\right)\\ &=\frac{2.5Q}{4\pi\epsilon_0}\\ \end{align}$$

Now, for the ball to float

$$\begin{align}F_e&=F_g\\ \frac{2.5qQ}{4\pi\epsilon_0}&=mg\\ q&=\frac{4\pi\epsilon_0mg}{2.5Q}\\ q&\approx17\times10^{-3}m\\ \end{align}$$

In other words - a lot.

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  • $\begingroup$ I'm supposed to start off by finding V. Have to derive E from V... $\endgroup$ – PoweredByOrange Feb 5 '15 at 6:23

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