Question

How does electricity flux, electric field enclosing a given charge vary when area enclosed by charge is doubled?

Answer 1

Lets imagine that the closed surface is a sphere of radius R. To double the area of the sphere’s surface, we must increase the radius by a factor of sqrt(2). This means the electric field at the surface of the sphere decreases by a factor of 2. The electric flux density vector D also decreases by a factor of 2. However the total electric flux out of the sphere is unchanged because the area ha s doubled. This is as Gauss’s law tells us, since the total flux out of a closed surface is equal to the charge inside the surface which hasn’t changed. HOPE THIS HELPS

Answer 2

If the area is doubled, the electric flux will also gets doubled.

Explanation:

The electric flux is defined as the number of electric field lines passing through a particular cross section. Its formula is given by :

$\phi=EA\ cos\theta$

E is the electric field

A is the are of cross section

When area enclosed by the charge is doubled, A' = 2A. New electric flux is given by :

$\phi'=EA'$

$\phi'=E(2A)$

$\phi'=2EA$

$\phi'=2\times \phi$

So, if the area is doubled, the electric flux will also gets doubled. Hence, this is the required solution.

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