A charge Q is enclosed by a gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
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.
To find the change in the electric flux when the radius is doubled, we need to consider the relationship between the electric flux and the radius.
The electric flux through a Gaussian surface is given by the formula:
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.
When the radius of the Gaussian surface is doubled, the surface area of the surface also doubles. This means that the electric flux is distributed over a larger surface area.
Mathematically, the surface area of a sphere is given by:
A = 4πR²
Where A is the surface area and R is the radius.
When the radius is doubled, the new surface area becomes:
A' = 4π(2R)² = 16πR²
Since the surface area has increased by a factor of 16, the electric flux will be spread out over a larger area. As a result, the outward electric flux will decrease.
Therefore, the outward electric flux will decrease when the radius is doubled.
To determine how the outward electric flux changes when the radius of a Gaussian spherical surface is doubled, we need to consider Gauss's Law.
Gauss's Law states that the total electric flux passing through a closed surface is directly proportional to the charge enclosed within that surface. Mathematically, it can be expressed as:
Φ = Q / ε₀
Where:
Φ is the electric flux,
Q is the charge enclosed, and
ε₀ is the permittivity of free space.
In this case, we have a charge Q enclosed by a Gaussian spherical surface of radius R. If we then double the radius to 2R, the question is asking whether the outward electric flux will increase, decrease, or remain the same.
To answer this question, we need to consider the relationship between the surface area of the Gaussian surface and the charge enclosed within it. The surface area of a sphere is given by:
A = 4πR²
If we double the radius to 2R, the new surface area becomes:
A' = 4π(2R)² = 4π(4R²) = 16πR²
Comparing the surface areas before and after doubling the radius, we see that the new surface area is four times larger than the original surface area. This means that the flux density (flux per unit area) of the Gaussian surface is reduced by a factor of 1/4.
Now, let's consider Gauss's Law again. The total electric flux passing through the Gaussian surface is proportional to the charge enclosed. If the charge Q remains the same when the radius is doubled, the outward electric flux will also reduce by a factor of 1/4.
In other words, when the radius is doubled, the outward electric flux will decrease by a factor of 1/4 or 25%.