Energy Density in the Electric Field of a Charged Sphere

AI Thread Summary
To calculate the energy density in the electric field near the surface of a charged sphere with a diameter of 10 cm and a potential of 8000V, the relevant equations include u=1/2(εE^2) and E=kq/r^2. The user initially attempted to find the charge (q) using capacitance (C) and the relationship q=CV, but encountered discrepancies in their calculations. After clarification, it was noted that confusion between diameter and radius led to an incorrect value for energy density. Correcting this error resulted in the accurate energy density of 0.11 J/m^3, aligning with the book's answer.
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Homework Statement


A charged isolated metal sphere of diameter 10cm has a potential of 8000V relative to V=0 at infinity. Calculate the energy density in the electric field near the surface of the sphere


Homework Equations


u=1/2\(epsilon x E^2)<br /> E=kq/r^2<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I have tried this like an example given in my book, which gives the q value, but since q=CV, can&#039;t i find C of the sphere, solve for q, put that into E, and subsequently solve for u? The answer in my book is .11 J/m^3, but when i use the above strategy, I get around .028. Could someone point me in the right direction?
 
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Express V in terms of q and r, then solve for q. It's a slightly roundabout way to calculate the energy density this way, but since you want to find q, this will work.
 
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I tried that using V=kq/r and put q into E=kq/r^2. I also got E using V=Ed and got about 80000 V/m or N/C both times. I still get around .028 J/m^3 for my answer when i put that into the density formula.
 
Oh, you probably just confused diameter and radius.
 
Thats exactly what i did wrong. I corrected and got .11J/m^3. thanks
 
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