Question

The flux associated with a spherical surface is $104\,N{m^2}{C^{ - 1}}$. What will be the flux if the radius of the spherical shell is doubled?

Answer 1

Hint:Electric flux is the measurement of the electric field passing through a given surface in electromagnetism, despite the fact that an electric field cannot flow by itself. It's a way of expressing the power of an electric field at any distance from the charge that's causing it.

Complete step by step answer:
If there is no charge inside a closed surface where an electric field line will end, any electric field line that enters the surface at one point must inevitably exit at another. As a result, the electric flux through a closed surface is zero if no charges exist within the enclosed volume.

Now assume that a charge q is present inside the spherical surface. Then the electric field will be generated around the spherical sphere and imaginary electric field lines will be there. Suppose the radius of the spherical sphere is ‘r’. Then the net electric flux over the spherical sphere is given by $\dfrac{q}{{{ \in _0}}}$.

So, we can observe from the above formula that the electric flux over any sphere is directly proportional to the charge present inside it. Now, according to the question the electric flux associated with the sphere is $104\,N{m^2}{C^{ - 1}}$. Now, if we double the radius of the sphere, no change will be there as we can see that the electric flux does not depend on the radius of the sphere.

Hence, there will be no change in the flux if the radius of the spherical shell is doubled.

Note: Electric field lines are merely a visual representation of field intensity and direction; they have no physical significance. The electric field strength, also known as the electric flux density: the number of "lines" per unit area, is proportional to the density of these lines.