Does electric flux depend on radius of Gaussian surface?

The electric flux density does not depend on the shape, size, and position of the closed surface concerning the charge placed inside. It only depends on the Gaussian surface. When changing the radius the electric flux through the surface remains the same. And the net charge enclosed the surface is also the same.

How does electric flux affected when radius is increased?

Since, on increasing the radius of the Gaussian surface, charge q remains unchanged, the flux through the spherical Gaussian surface will not be affected when its radius is increased.

What happens to electric flux when area is doubled?

If the area is doubled, the electric flux will also gets doubled. When area enclosed by the charge is doubled, A’ = 2A. New electric flux is given by : So, if the area is doubled, the electric flux will also gets doubled.

What is the relationship between electric flux and size of size of Gaussian surface?

The flux Φ of the electric field →E through any closed surface S (a Gaussian surface) is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ϵ0): Φ=∮S→E⋅ˆndA=qencϵ0.

Does radius affect electric flux?

Gauss’s Law – The total electric flux through any closed surface is proportional to the total electric charge inside the surface. Point Charge Inside a Spherical Surface: – The flux is independent of the radius R of the sphere.

What is the effect of change in radius of Gaussian surface on the flux?

As ϕ=q∈0, so flux does not depend upon the radius of Gaussion surface, it will remain unchanged.

How does electric flux due to point charge enclosed by a Gaussian surface get affected when its radius is increased?

How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? That is, on increasing the radius of the gaussian surface, charge q remains unchanged. So, flux through the gaussian surface will not be affected when its radius is increased.

How does the electric flux?

electric flux, property of an electric field that may be thought of as the number of electric lines of force (or electric field lines) that intersect a given area. The negative flux just equals in magnitude the positive flux, so that the net, or total, electric flux is zero.

Which does not affect electric flux?

It is important to note that while the electric flux is not affected by charges that are not within the closed surface, the net electric field, E, in the Gauss’ Law equation, can be affected by charges that lie outside the closed surface. Thus, the SI base units of electric flux are kg·m3·s−3·A−1.

When the radius of Gaussian surface is doubled electric flux remains same?

Even if the radius is doubled, then the outward electric flux will remain the same as \[\dfrac{q}{{{\varepsilon _0}}}\]. This is because the outward electric flux is independent of the distribution of the charges and the separation between them inside the closed surface. Hence, option D is the correct answer.

How does the electric flux change if the radius of the Gaussian surface is triple?

What happens to the electric flux through a Gaussian surface?

A spherical gaussian surface surrounds a point charge $q$. Describe what happens to the: flux through the surface if 2) The radius of the sphere is doubled Our teacher said the electric flux will not change 3) The shape of the surface is changed to that of a cube

How does the electric flux get affected when the radius is increased?

How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? Please log in or register to add a comment. Here, q is the charge enclosed by the gaussian surface. That is, on increasing the radius of the gaussian surface, charge q remains unchanged.

What happens when the radius of a Gaussian surface is doubled?

If the radius of the spherical Gaussian surface is doubled, the surface area will increase by 4 times. On the other hand, the electric field at all​ points on the bigger surface will decrease by a factor of 4. Hence the flux through the surface remains the same.

Why electric charge does not enter inside a closed Gaussian surface?

I think the correct answer for your question is that the electric charge outside of a closed Gaussian surface does not enter inside because the surface is enclosed. Take a look at the Faraday’s cage and you can also take into account why we are not affected when driving in a car and get struck by a lightening.