How to Find the Potential Energy of a Cloud?

AI Thread Summary
To find the potential energy of a cloud, the maximum electric field is given as 3x10^6 Vm^-1, with a distance of 1000 meters to Earth. The potential difference can be calculated using the formula V=Ed, where E is the electric field and d is the distance. The discussion highlights that the second formula for energy stored in an electric field is unnecessary for this problem, as the electric field value is already provided. It is emphasized that the potential difference can be directly calculated without needing to derive the electric field again. The solution approach is clarified, focusing on the correct application of the given equations.
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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


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


The Attempt at a Solution



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

using V=Ed
E = 8.23\times10^8 \times 1000
=8.23x10^11 volts


is this right?
 
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boneill3 said:

Homework Equations


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

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:
\mu = \frac{1}{2}(\epsilon_{0}\times E^2)
3\times 10^6 = \frac{1}{2}(8.85\times 10^-12\times E^2)

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.
 
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