Question

Suppose you have a sphere of radius 1.00 m that carries a charge of +1.00 µC at its center. The radius of the sphere is changed to 0.500 m. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere? The flux and field both increase. The flux increases and the field decreases. The flux decreases and the field remains the same. The flux decreases and the field increases. The flux remains the same and the field increases. The flux and field both decrease.

Answer 1

Answer:

The flux remains the same and the field increases.

Explanation:

According to the Gauss' law, the surface integral of the electric field over a closed surface, called the Gaussian surface, is equal to $\dfrac{1}{\epsilon_o}$ times the net charge $q$ enclosed by the surface.

$\oint \vec E \cdot d\vec A=\dfrac{q}{\epsilon_o}\ \ \ \ ............\ (1).$

where,

$\epsilon_o$ = electrical permititivity of free space.

The term on the LHS of Gauss law' $\oint \vec E \cdot d\vec A$ is the electric flux $\phi$ through the Gaussian surface.

Therefore,

$\phi=\dfrac{q}{\epsilon_o}\ \ \ .....................\ (2).$

From equation (2), the electric flux through the sphere depends only on the charge enclosed by the sphere and not on the size of the sphere, therefore, when the radius of the sphere is changed to 0.500 m from 1.00 m, the electric flux through the surface will not change.

Consider the Gaussian sphere of radius $r$ same as that of the given sphere and concentric with the given sphere, then,

$\oint \vec E \cdot d\vec A=\oint E\ dA$

This is because the direction of the electric field through a Gaussian spherical surface and its surface area element $d\vec A$ is always normal to it.

Therefore,

$\oint \vec E \cdot d\vec A = E\oint dA=E\ 4\pi r^2$

Using equation (1),

$E\ 4\pi r^2=\dfrac{q}{\epsilon_o}\\E=\dfrac{q}{4\pi r^2 \epsilon_o}\\E\propto \dfrac{q}{r^2}$

The electric field is inversely proportional to the square of the radius of the sphere, therefore, on decreasing the radius from  1.00 m to 0.500 m, the electric field increases.

Thus, the correct answer is "The flux remains the same and the field increases".

Answer 2

When radius of the sphere is changed to 0.500 m, the flux remains the same and the field increases.

Electric flux

Since electric flux through a Gaussian surface A is Ф = q/ε₀. where

• q = charge and
• ε = permittivity of free- space

Since the electric flux depends on the charge only, when the radius is reduced from 1.00 m to 0.500 m, the electric flux remains the same since the charge does not change.

Electric field

Also, the electric flux, Ф = ∫E.dA = q/ε where

• E = electric field,
• A = area of the Gaussian surface,
• q = charge and
• ε = permittivity of free- space

Now, since the area is spherical, and E is in the direction of the normal to the surface, ∫E.dA = ∫EdA

= E∫dA

= E4πr².

E4πr² = q/ε

E = q/4πεr²

Since q and ε are constant,

E ∝ 1/r²

So, when the radius decreases from 1.00 m to 0.500 m, the electric field increases.

Thus, the electric flux remains the same, but the electric field increases.

So, when radius of the sphere is changed to 0.500 m, the flux remains the same and the field increases.

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