A sphere of radius 1cm has a potential of 8000V. Then the energy density near its surface will be.......(in MKS).

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31.87

63.43

8.52

2.83

answer is D.

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Detailed Solution

Concept: A sphere is given having compass = R = 1 cm

https://www.vedantu.com/question-sets/b226311d-992a-4bae-bd95-bb3852506c5e8460150291789016928.png

The electric eventuality at the face of sphere = 8000V
We know that the relation between electric field and electric eventuality is given as
E = VR
Putting the value of ‘ V ’ and ‘ R ’ in below equation, we get
E = 80001V/ cm = 8000V/ cm = 8 × 105V/ m
thus the electric field at face of sphere = 8000V/ cm = 8 × 105V/ m
Now, we know that the formula for calculating the energy viscosity of electric field is given as
U = 1/2 ∈0E2, where
∈ 0 is permittivity of free space = 8.85 × 10 − 12 m − 3 kg − 1s4A2 E = electric field at face of sphere = 8 × 105V/ m
Using the values in the equation above, we obtain
U = 1/2 ×8.85 × 10 − 12 ×( 8 × 105) 2 = 8.85 × 32 × 10 − 2 = 2.832 J/ Kgs.
Hence, the correct option is 4.

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