91Ó°ÊÓ

• The electric flux =$-750N.m^2/C$
• Gaussian surface radius = 10.0 cm

Gauss law describes the relation between charge and electric field in a static situation. The equation for Gauss law is,

${\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{âˆ\dots }\mathbf{=}{\mathbf{q}}_{\mathbf{enc}}$

Here, ${\mathrm{q}}_{\mathrm{enc}}$is the net charge inside an imaginary closed surface and$\mathbf{âˆ\dots }$ is the net flux of the electric field through the surface.

By doubling the area, the only surface has increased. It does not change the amount of charge enclosed by the surface area. From the equation,

${\mathrm{Î\mu }}_0\mathrm{Φ}=q_{enc}$

We can see that flux depends on the enclosed charge. Therefore, the flux will remain the same as$-750N.m^2/C$ .

We use $\mathrm{Ï•}=\mathrm{q}/{\mathrm{Î\mu }}_0$to calculate the charge as:

$$\begin{gathered} q=âˆ\dots {\mathrm{Î\mu }}_0 \\ =\left(8.85×{10}^{-12}{C}^2/N.m^2\right)\left(-750N.m^2/C\right) \\ =-6.64×{10}^{-9}C \end{gathered}$$

Therefore, the charge on the particle is $-6.64×{10}^{-9}C$.