The energy density of air medium is
44.25 x 10 J/m. The intensity of the electric
field in the medium is
(A) 300 N/C (B) 3 N/C
(C) 305 N/C (D) 316.2 N/C
() 300 NC
3NENC
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hi,
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Que: The energy density of air medium is 44.25 x 10^8 J/m. The intensity of the electric field in the medium is 316.2N/C.
Solution:
The relation between Energy Density UE and electric field intensity E is given as
UE = 1/2ε0E2 (1)
From which the E can be obtained as
E = √2UE / ε0−−− (2)
Where ε0 = 8.8541 x 10^-12 F/m and UE is given as 44. 25 x 10^-8 J/m3
Then E is obtained as , E = 316. 15 N/C.
_______
Que: The energy density of air medium is 44.25 x 10^8 J/m. The intensity of the electric field in the medium is 316.2N/C.
Solution:
The relation between Energy Density UE and electric field intensity E is given as
UE = 1/2ε0E2 (1)
From which the E can be obtained as
E = √2UE / ε0−−− (2)
Where ε0 = 8.8541 x 10^-12 F/m and UE is given as 44. 25 x 10^-8 J/m3
Then E is obtained as , E = 316. 15 N/C.
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