Looking for classes? Ksquare Career Institute, Bengaluru →
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
A
Be doubled
B
Increase four times
C
Be reduced to half
D
Remain the same
Detailed Solution
By Gauss's law, the total outward electric flux through a closed surface is $\phi_E = \frac{Q_{enclosed}}{\varepsilon_0}$
The flux depends only on the charge enclosed, not on the size or shape of the surface.
When the radius is doubled, the charge enclosed is still Q.
(The field at the surface becomes one-fourth while the area becomes four times, so their product is unchanged.)
Hence the outward electric flux remains the same.
The flux depends only on the charge enclosed, not on the size or shape of the surface.
When the radius is doubled, the charge enclosed is still Q.
(The field at the surface becomes one-fourth while the area becomes four times, so their product is unchanged.)
Hence the outward electric flux remains the same.