A spherical Gaussian surface encloses a charge of 8.85 1010 C. (i) Calculate the
electric flux passing through the surface (ii) How would the flux change if the radius
of the Gaussian surface is doubled and why?
Answers
Answered by
69
using Gauss law: integral E . dS = q/ε
Flux = 8.85 * 10^-10 /ε units.
If the radius is doubled for the spherical surface still the flux remains same, as that depends only on the charge enclosed..... in other words, electric field reduces and radius increases, E * R^2 remains same.
Flux = 8.85 * 10^-10 /ε units.
If the radius is doubled for the spherical surface still the flux remains same, as that depends only on the charge enclosed..... in other words, electric field reduces and radius increases, E * R^2 remains same.
Answered by
43
Answer:
(i)
we know that the flux linked with a Gaussian surface is given as
Φ = E.ds = q/ε0
here,
q = 8.85x10-8 C
ε0 = 8.85x10-12 F/m
so,
Φ = 8.85x10-8 / 8.85x10-12
thus, flux will be
Φ = 104 Nm2/C
.
(ii)
The flux does not depend upon the size of the Gaussian surface but only on the charges enclosed. So, in this case also
Φ = 104 Nm2/C.
Explanation:
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