Solution

Rest frame S'

In S', the sphere is at rest, and the charge is uniformly distributed:

$\sigma' = \frac{Q}{4\pi R^2} \quad (\text{uniform surface density})$.

The electric field just outside the surface is radial and is (in CGS units):

$E' = \frac{Q}{R^2}$

There is no magnetic field in S': $\mathbf{B}' = 0$.

Transformation to the laboratory frame S

The laboratory frame S moves relative to S' with velocity $\mathbf{v} = -v\hat{\mathbf{z}}$
Equivalently, the sphere moves in S with velocity $\mathbf{v} = v\hat{\mathbf{z}} (v = \beta c)$.

The components of the electric field parallel and perpendicular to $\mathbf{v}$ transform as:

$E_\parallel = E'_\parallel, \qquad$
$\mathbf{E}_\perp = \gamma (\mathbf{E}'_\perp + \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}'_\perp$

with $\gamma = 1/\sqrt{1-\beta^2}$

Electric field on the surface (in S)

In S', the electric field is radial:

At the equator $(\theta' = \pi/2): \mathbf{E}'$ is perpendicular to $\mathbf{v} → \mathbf{E}'_\perp = \mathbf{E}'$

At the poles $(\theta' = 0): \mathbf{E}'$
is parallel to $\mathbf{v} → \mathbf{E}'_\parallel = \mathbf{E}'$

In S:

Equator:$E = \gamma E' = \dfrac{\gamma Q}{R^2} → maximum$

Poles:$E = E' = \dfrac{Q}{R^2} → minimum$

Thus:

$\boxed{E_{\max} = \frac{Q}{R^2\sqrt{1-\beta^2}}}, \qquad$

$E_{\min} = \frac{Q}{R^2}$

Surface charge density (in S)

The total charge Q is an invariant. In S, the sphere appears as an ellipsoid with semi-axes R, R, $R\sqrt{1-\beta^2}$
(contracted in the direction of motion).
The surface density $\sigma$ is obtained from the transformation of the four-current or directly from charge conservation when transforming area elements.

Equator (normal perpendicular to $\mathbf{v}): \sigma_{\max} = \gamma \sigma'$
· Poles (normal parallel to $\mathbf{v}): \sigma_{\min} = \sigma'/\gamma$

Therefore
$\sigma_{\max} = \frac{Q}{4\pi R^2\sqrt{1-\beta^2}}, \qquad$

$\sigma_{\min} = \frac{Q\sqrt{1-\beta^2}}{4\pi R^2}$