A charge Q is enclosed by a gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

1 answer

The outward electric flux through the gaussian surface is given by the electric flux equation:

Φ = Q / ε₀

where Φ is the electric flux, Q is the charge enclosed by the gaussian surface, and ε₀ is the permittivity of free space.

If the radius of the gaussian surface is doubled (i.e., becomes 2R), then the surface area of the gaussian surface also doubles, since the surface area of a sphere is proportional to the square of the radius:

A = 4πR²

When the radius is doubled to 2R, the surface area becomes:

A' = 4π(2R)² = 4π(4R²) = 16πR² = 16A

Therefore, the new surface area is 16 times the original surface area.

Since the charge enclosed by the gaussian surface remains the same (Q), the electric flux through the gaussian surface will also increase by a factor of 16:

Φ' = Q / ε₀

Φ' = 16Q / ε₀

Therefore, the outward electric flux will be 16 times the original electric flux.