wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

A sphere of radius 1cm has potential of 8000 V ,then what will be energy density near its surface in J/m^3?

Open in App
Solution

Answer: (B) 2.83 (J/m3)

Energy density = (1/2)∈0E2

= (1/2) × 8.86 × 10–12 × (v/r)2

= 4.43 × 10–12 × [(8000) / (10–2)]2

= 283.5 × 10–2

= 2.83 J/m3 ​​​​​​

Hope this helps :)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Realistic Collisions
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon