The flight of the housefly

Electricity and Magnetism Level 4

A housefly is sitting between the plates of a parallel-plate capacitor. The plates of the capacitor are disks of radius $R=30 cm$ separated by a distance $d=1 cm$ . The distance between the fly and the positive plate of the capacitor is $r_{0}=\frac{d}{2}(1-\epsilon)$ with $\epsilon=10^{-4}$ ( note that the fly is very close to the center of the capacitor). The housefly decides to exit the capacitor by flying along an equipotential line. What is the maximum distance in meters from the center of the capacitor the housefly reaches along the equipotential line?

The answer is 21.2.

1 solution

Tran Dinh Duy Vu
Sep 8, 2013

When the house fly is in the capacitor, because the field is uniform and it is obvious that the potential in the middle of the capacitor is zero, the potential of the fly is : $V= E.\frac{d\epsilon}{2} = \frac{Q}{\pi R^{2} \varepsilon}.\frac{d\epsilon}{2}$

since $\epsilon$ is very small, the potential is very small too. This potential can be obtained again when the fly is very far from the capacitor (the potential at that point is nearly zero). Hence, the capacitor cab be treated as a dipole with dipole momentum $p= Q.d$

The potential cause by a dipole is :

$V= \frac{p\sin \alpha}{ 4\pi \varepsilon r^{2}} = \frac{p}{4\pi \varepsilon r_{max}^{2}}$

The ultimate equation is :

$\frac{Q}{\pi R^{2} \varepsilon}.\frac{d\epsilon}{2} = \frac{p}{4\pi \varepsilon r_{max}^{2}} then r_{max} =\frac{R}{\sqrt{2\epsilon}}= 21.2m$

I tried to fix your formatting.

When the house fly is in the capacitor, because the field is uniform and it is obvious that the potential in the middle of the capacitor is zero, the potential of the fly is $$V= E\frac{d\epsilon}{2} = \frac{Q}{\pi R^{2} \varepsilon 0}\frac{d\epsilon}{2}.$$ Since $\epsilon$ is very small, the potential is very small too. This potential can be obtained again when the fly is very far from the capacitor (the potential at that point is nearly zero). Hence, the capacitor can be treated as a dipole with dipole moment $p= Qd$ . The potential cause by a dipole is $$V= \frac{p\sin \alpha}{ 4\pi \varepsilon 0 r^{2}} = \frac{p}{4\pi \varepsilon 0 r {max}^{2}}$$ The ultimate equation is: $$\frac{Q}{\pi R^{2} \varepsilon 0}.\frac{d\epsilon}{2} = \frac{p}{4\pi \varepsilon 0 r {max}^{2}},$$ then $$r {max} =\frac{R}{\sqrt{2\epsilon}}= 21.2 \ \mathrm{m}$$

Ricky Escobar - 7 years, 9 months ago

Not related to the solution but this problem is repeated for me. I remember I solved the problem in the past so I guess I shouldn't have received the problem.

Pranav Arora - 7 years, 9 months ago

is it legit to simply assume the capacitor is a dipole? I tried calculating the potential of each disk directly and got a slightly different answer

Jiahai Feng - 7 years, 9 months ago

Legen-wait for it-dary.

Priyatam Roy - 7 years, 8 months ago

The flight of the housefly

The answer is 21.2.

1 solution

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