How to Find the Potential Energy of a Cloud?

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boneill3
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Homework Statement



Assuming the maximum electric field sustained by dry air in a cloud is 3x10^6 Vm^-1
And a distanceof 1000 meters between Earth and cloud. The cloud is 4km long and 1 km wide. ausume uniform eletric field. Find the potential difference.

Homework Equations


[itex]V=Ed[/itex]
[itex]\mu = \frac{1}{2}(\epsilon_{0}\times E^2)[/itex]


The Attempt at a Solution



[itex]\mu = \frac{1}{2}(\epsilon_{0}\times E^2)[/itex]
[itex]3\times 10^6 = \frac{1}{2}(8.85\times 10^-12\times E^2)[/itex]
[itex]E=\sqrt{\frac{3\times 10^6}{\frac{1}{2}(8.85\times10^-12\timesE^2)}}[/itex]
[itex]E = 8.23\times10^8[/itex]

using [itex]V=Ed[/itex]
[itex]E = 8.23\times10^8 \times 1000[/itex]
=8.23x10^11 volts


is this right?
 
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boneill3 said:

Homework Equations


[itex]V=Ed[/itex]
[itex]\mu = \frac{1}{2}(\epsilon_{0}\times E^2)[/itex]

You don't need the second formula (which is for the energy stored in an electric field, I believe. The first equation, along with the given information, should be enough to solve the problem.

boneill3 said:
[itex]\mu = \frac{1}{2}(\epsilon_{0}\times E^2)[/itex]
[itex]3\times 10^6 = \frac{1}{2}(8.85\times 10^-12\times E^2)[/itex]

Why have you plugged in the value of the electric field, E, for mu? The electric field, E, is already given to you. You don't need to solve for it.