Question

A charged particle causes an electric flux of -750 N.m²/Cto pass through a spherical Gaussian surface of 10.0 cmradius centered on the charge.

(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

(b) What is the charge of the particle?

Short answer

a) If the radius of the Gaussian surface were doubled, the amount of flux passing through the surface is$-750\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$ .

b) Charge on the particle is $-6.64\times {10}^{9}C$.

Step-by-step solution

Listing the given quantities

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

Understanding the concept of Gauss law

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

${\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\varnothing }\mathbf{=}{\mathbf{q}}_{\mathbf{enc}}$

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

(a) Calculation of the flux passing 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,

${\epsilon }_0\Phi =q_{enc}$

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

(b) Calculation of the charge on the particle

We use $\mathrm{\varphi }=\mathrm{q}/{\mathrm{\epsilon }}_0$to calculate the charge as:

$$\begin{gathered} q=\varnothing {\epsilon }_0 \\ =\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\left(-750N.m^2/C\right) \\ =-6.64\times {10}^{-9}C \end{gathered}$$

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